A parallel plate capacitor of area and separation is charged initially to . If the medium between the plate gets slightly conducting and the plate loses the charge initially at the rate of , then what is the magnetic field between the plates ?
1
2
3
4
zero
Official Solution
Correct Option:
(4)
Magnetic field between the plates in this case is zero.
02
PYQ 2013
hard
physicsID: jee-main
Choose the correct sketch of the magnetic field lines of a circular current loop shown by the dot ? and the cross ?.
1
2
3
4
Official Solution
Correct Option:
(1)
If magnetic field is perpendicular and into the plane of the paper, it is represented by cross and if the direction of the magnetic field is perpendicular out of the plane of the paper it is represented by dot ?.
03
PYQ 2013
medium
physicsID: jee-main
This question has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two Statements. Higher the range, greater is the resistance of ammeter. To increase the range of ammeter, additional shunt needs to be used across it.
1
Statement - I is true, Statement - II is true, Statement - II is the correct explanation of Statement-I
2
Statement - I is true, Statement - II is true, Statement - II is not the correct explanation of Statement-I
3
Statement - I is true, Statement - II is false
4
Statement - I is false, Statement - II is true
Official Solution
Correct Option:
(4)
Answer (d) Statement - I is false, Statement - II is true
04
PYQ 2014
easy
physicsID: jee-main
A conductor lies along the -axis at and carries a fixed current of in direction (see figure). For a field , find the power required to move the conductor at constant speed to in . Assume parallel motion along the -axis
1
1.57 W
2
2.97 W
3
14.85 W
4
29.7 W
Official Solution
Correct Option:
(2)
Average Power
05
PYQ 2015
medium
physicsID: jee-main
Two coaxial solenoids of different radii carry current in the same direction. Let be the magnetic force on the inner solenoid due to the outer one and be the magnetic force on the outer solenoid due to the inner one. Then,
1
is radially outwards and = 0
2
is radially inwards and is radially outwards
3
is radially inwards and = 0
4
Official Solution
Correct Option:
(4)
Cross-sectional view (Both solenoids are taken to be ideal in nature.) Both wires will attract each other, but net force on each wire will be zero. Concept:
Two current carrying elements attract each other if direction of current is same.
06
PYQ 2015
medium
physicsID: jee-main
A proton (mass m) accelerated by a potential difference flies through a uniform transverse magnetic field . The field occupies a region of space by width . If be the angle of deviation of proton from initial direction of motion (see figure), the value of will be :
1
2
3
4
Official Solution
Correct Option:
(3)
07
PYQ 2016
easy
physicsID: jee-main
A galvanometer having a coil resistance of gives a full scale deflection, when a current of is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of is :
1
2
3
4
Official Solution
Correct Option:
(1)
Maximum voltage that can be applied across the galvanometer coil If is the shunt resistance :
08
PYQ 2016
medium
physicsID: jee-main
To know the resistance of a galvanometer by half deflection method, a battery of emf and resistance is used to deflect the galvanometer by angle . If a shunt of resistance is needed to get half deflection then and are related by the equation :
1
2
3
4
Official Solution
Correct Option:
(2)
09
PYQ 2017
medium
physicsID: jee-main
When a current of is passed through a galvanometer having a coil of resistance , it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range is :
1
2
3
4
Official Solution
Correct Option:
(1)
Let series resistance be R.
10
PYQ 2018
medium
physicsID: jee-main
An electron , a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii respectively in a uniform magnetic field . The relation between is :
1
2
3
4
Official Solution
Correct Option:
(2)
Mass of electron is least and charge So,
11
PYQ 2018
medium
physicsID: jee-main
In a circuit for finding the resistance of a galvanometer by half deflection method, a battery and a high resistance of are used. The figure of merit of the galvanometer is division. In the absence of shunt resistance, the galvanometer produces a deflection of divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of , is closest to :
1
2
3
4
Official Solution
Correct Option:
(4)
Current required by unit deflection is For, current is
Let is resistance of galvanometer. Then,
Also in half deflection method,
12
PYQ 2019
medium
physicsID: jee-main
A particle having the same charge as of electron moves in a circular path of radius under the influence of a magnetic field of . If an electric field of makes it to move in a straight path, then the mass of the particle is (Given charge of electron =
1
2
3
4
Official Solution
Correct Option:
(1)
....(i) Path is straight line it ....(ii) From equation (i) & (ii)
13
PYQ 2019
medium
physicsID: jee-main
A galvanometer, whose resistance is , has divisions in it. When a current of A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range , it should be connected to a resistance of:
1
6250 ohm
2
250 ohm
3
200 ohm
4
6200 ohm
Official Solution
Correct Option:
(3)
14
PYQ 2019
medium
physicsID: jee-main
A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of . The magnetic field at point will be close to :
1
2
3
4
Official Solution
Correct Option:
(1)
15
PYQ 2019
easy
physicsID: jee-main
Two very long, straight, and insulated wires are kept at angle from each other in - plane as shown in the figure. These wires carry currents of equal magnitude , whose directions are shown in the figure. The net magnetic field at point will be :
1
2
3
4
Official Solution
Correct Option:
(1)
Magnetic field at point
16
PYQ 2019
medium
physicsID: jee-main
A circular coil having turns and radius carries a current . It is held in the plane in a magnetic field . The torque on the coil due to the magnetic field is :
1
2
3
Zero
4
Official Solution
Correct Option:
(1)
Magnetic moment of coil =
Torque on loop (coil) =
17
PYQ 2019
medium
physicsID: jee-main
One of the two identical conducting wires of length is bent in the form of a circular loop and the other one into a circular coil of identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop to that at the centre of the coil , i.e., will be
1
2
3
4
Official Solution
Correct Option:
(3)
18
PYQ 2019
medium
physicsID: jee-main
Find the magnetic field at point due to a straight line segment of length carrying a current of . (See figure) (
1
2
3
4
Official Solution
Correct Option:
(4)
d = 4 cm
19
PYQ 2019
medium
physicsID: jee-main
The magnetic moment of an octahedral homoleptic Mn(II) complex is . The suitable ligand for this complex is :
1
2
3
4
ethylenediamine
Official Solution
Correct Option:
(2)
(no of unpaired.e-) = 5 Cation confn only possible for relatively weak ligand.
20
PYQ 2019
medium
physicsID: jee-main
An electron, moving along the x-axis with an initial energy of , enters a region of magnetic field at (See figure). The field extends between and . The electron is detected at the point on a screen placed away from the point . The distance between and (on the screen) is : (electron's charge = mass of electron =
1
12.87 cm
2
1.22 cm
3
11.65 cm
4
2.25 cm
Official Solution
Correct Option:
(1)
cm
=
21
PYQ 2019
medium
physicsID: jee-main
Two wires & are carrying currents & as shown in the figure. The separation between them is . A third wire carrying a current is to be kept parallel to them at a distance from A such that the net force acting on it is zero. The possible values of are :
1
and
2
3
and
4
and
Official Solution
Correct Option:
(2)
Net force on wire carrying current I per unit length is
22
PYQ 2019
medium
physicsID: jee-main
A proton and an -particle (with their masses in the ratio of and charges in the ratio of ) are accelerated from rest through a potential difference . If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii of the circular paths described by them will be :
1
2
3
4
Official Solution
Correct Option:
(1)
The correct option is(A):
23
PYQ 2020
easy
physicsID: jee-main
A galvanometer of resistance is converted into a voltmeter of range by connecting a resistance in series with it. The additional resistance that should be connected in series with to increase the range of the voltmeter to will be :
1
2
3
4
Official Solution
Correct Option:
(2)
(1)
24
PYQ 2021
medium
physicsID: jee-main
A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT. The torque acting on the coil when a current of 0.2 A is passed through it and its plane becomes parallel to the magnetic field will be Nm. The value of x is __________.
Official Solution
Correct Option:
(1)
Step 1: Understanding the Question:
We need to find the torque on a triangular current-carrying coil placed in a magnetic field. The orientation is specified, which allows us to find the maximum torque. Step 2: Key Formula or Approach:
1. The torque ( ) on a current loop in a uniform magnetic field (B) is given by , where is the magnetic dipole moment of the loop.
2. The magnitude of the torque is , where is the angle between and .
3. The magnetic moment is given by , where N is the number of turns (here N=1), I is the current, and A is the area of the loop.
4. Torque is maximum when , which occurs when is perpendicular to . The direction of is normal to the plane of the coil. If the plane of the coil is parallel to , then its normal is perpendicular to , so . Step 3: Detailed Explanation:
Given values:
Side of equilateral triangle, a = 10 cm = 0.1 m
Magnetic field, B = 20 mT = 20 10 T
Current, I = 0.2 A
Number of turns, N = 1 First, calculate the area (A) of the equilateral triangle coil:
The plane of the coil is parallel to the magnetic field, so the angle between the magnetic moment (normal to the plane) and the magnetic field is 90 . Thus, the torque is maximum.
Substitute the values:
The problem states that the torque is Nm.
Comparing our result with the given expression: Step 4: Final Answer:
The value of x is 3.
25
PYQ 2021
medium
physicsID: jee-main
The fractional change in the magnetic field intensity at a distance 'r' from centre on the axis of current carrying coil of radius 'a' to the magnetic field intensity at the centre of the same coil is : (Take )
1
2
3
4
Official Solution
Correct Option:
(4)
Step 1: Understanding the Concept: The magnetic field due to a circular coil at an axial point is compared with the field at the center. For points close to the center, we use a binomial approximation. Step 2: Key Formula or Approach: Magnetic field at axial distance : . Field at center ( ): . Step 3: Detailed Explanation: Express in terms of :
Since , apply binomial expansion :
Fractional change:
Step 4: Final Answer: The fractional change is .
26
PYQ 2021
medium
physicsID: jee-main
A proton, a deuteron and an particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them is __________ and their speed is __________, in the ratio.
1
4 : 2 : 1 and 2 : 1 : 1
2
2 : 1 : 1 and 4 : 2 : 1
3
1 : 2 : 4 and 1 : 1 : 2
4
1 : 2 : 4 and 2 : 1 : 1
Official Solution
Correct Option:
(2)
Step 1: Force . Since momentum is same, .
Step 2: . Since are constant, .
Step 3: Ratio of : Proton ( ), Deuteron ( ), ( ).
Ratio : .
Step 4: Ratio of speeds ( ):
Proton ( ), Deuteron ( ), ( ).
Ratio : .
27
PYQ 2021
medium
physicsID: jee-main
If the maximum value of accelerating potential provided by a radio frequency oscillator is 12 kV. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is __________. [m = 1.67 10 kg, e = 1.6 10 C, Speed of light = 3 10 m/s]
Official Solution
Correct Option:
(1)
Step 1: Understanding the Question:
A proton in a cyclotron gains energy each time it crosses the gap between the dees. We are given the accelerating potential and the final desired speed. We need to find how many full revolutions it takes to reach that speed. Step 2: Key Formula or Approach:
1. Calculate the final kinetic energy (K ) of the proton using .
2. In a cyclotron, the particle is accelerated twice in each revolution. The energy gained in one revolution is , where is the accelerating potential.
3. The total number of revolutions (N) is the total kinetic energy gained divided by the energy gained per revolution: . Step 3: Detailed Explanation:
Given values:
Accelerating potential, V = 12 kV = 12 10 V
Proton mass, m = 1.67 10 kg
Proton charge, e = 1.6 10 C
Speed of light, c = 3 10 m/s
Final speed, v = c/6 = (3 10 ) / 6 = 0.5 10 m/s First, calculate the final kinetic energy (K ) in Joules:
Next, calculate the energy gained per revolution in Joules:
Finally, calculate the number of revolutions (N):
The number of revolutions must be an integer. Since the question asks for the number of revolutions to achieve the speed, we take the integer part. Step 4: Final Answer:
The number of revolutions made by the proton is 543.
28
PYQ 2021
medium
physicsID: jee-main
A proton and an -particle enter into a magnetic field at right angles. The ratio of the radii of trajectory of proton to that of -particle is 2 : 1. The ratio of is:
1
1 : 4
2
4 : 1
3
8 : 1
4
1 : 8
Official Solution
Correct Option:
(2)
Step 1: The formula for radius in a magnetic field is .
Step 2: Express Kinetic Energy : (since is constant).
Step 3: Let proton have and -particle have .
Step 4: Calculate the ratio:
The ratio is 4 : 1.
29
PYQ 2022
medium
physicsID: jee-main
Given below are two statements:
Statement I: The electric force changes the speed of the charged particle and hence changes its kinetic energy; whereas the magnetic force does not change the kinetic energy of the charged particle. Statement II: The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field.
In the light of the above statements, choose the most appropriate answer from the options given below:
1
Both statement I and statement II are correct
2
Both statement I and statement II are incorrect
3
Statement I is correct but statement II is incorrect
4
Statement I is incorrect but statement II is correct
Official Solution
Correct Option:
(3)
The correct answer is (C) : Statement I is correct but statement II is incorrect. Electric field accelerates the particle in the direction of field
and magnetic field accelerates the particle perpendicular to the field
30
PYQ 2022
medium
physicsID: jee-main
A charge particle is moving in a uniform magnetic field If it has an acceleration of , then the value of will be
1
3
2
6
3
12
4
2
Official Solution
Correct Option:
(2)
The correct option is (B) : 6 As
So , are ⊥ to each other Hence ,
31
PYQ 2022
medium
physicsID: jee-main
A bar magnetic having a magnetic moment of 2.0 \times 105 JT–1, is placed along the direction of uniform magnetic field of magnitude B = 14 \times 10–5 T. The work done in rotating the magnet slowly through 60° from the direction of field is:
1
14 J
2
8.4 J
3
4 J
4
1.4 J
Official Solution
Correct Option:
(1)
- .
–
The correct option is (A): 14 J
32
PYQ 2022
medium
physicsID: jee-main
A cyclotron is used to accelerate protons.If the operating magnetic field is 1.0 T and the radius of the cyclotron ‘dees’ is 60 cm,the kinetic energy of the accelerated protons in MeV will be
1
12
2
18
3
16
4
32
Official Solution
Correct Option:
(2)
The correct answer is (B) : 18
= 18MeV
33
PYQ 2022
medium
physicsID: jee-main
Two 10 cm long, straight wires, each carrying a current of 5 A are kept parallel to each other. If each wire experienced a force of 10–5 N, then separation between the wires is _____ cm.
Official Solution
Correct Option:
(1)
It should be mentioned, 10 cm wire is part of long wire. Force experienced by unit length of wire
Fig. Force Experienced
d = 5 cm
34
PYQ 2022
easy
physicsID: jee-main
A charge particle is moving in a uniform field . If it has an acceleration of , then the value of α will be
1
3
2
6
3
12
4
2
Official Solution
Correct Option:
(2)
As magnetic force is perpendicular to magnetic field. So, must be . So,
So, the correct option is (B): .
35
PYQ 2022
medium
physicsID: jee-main
The dimensions of will be :(if : permeability of free space andB : magnetic field)
1
2
3
4
Official Solution
Correct Option:
(3)
From the above correct answer is option (C):
36
PYQ 2022
medium
physicsID: jee-main
A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for deflection in the galvanometer. Which of the below is true for measuring value of G?
1
deflection method cannot be used for determining the resistance of the galvanometer.
2
deflection method can be used and in this case the G equals to twice the value of shunt resistance(s).
3
deflection method can be used and in this case, the G equals to three times the value of shunt resistance(s).
4
deflection method can be used and in this case the G value equals to the shunt resistance(s).
Official Solution
Correct Option:
(2)
The circuit for the given situation is:
Since G and S are in parallel,
So, G equals twice the value of shunt resistance.
Hence, the correct option is (B): deflection method can be used and in this case the G equals to twice the value of shunt resistance(s).
37
PYQ 2022
medium
physicsID: jee-main
A singly ionized magnesium atom (A=24) ion is accelerated to kinetic energy 5keV and is projected perpendicularly into a magnetic field B of the magnitude 0.5 T. The radius of path formed will be ________ cm.
Official Solution
Correct Option:
(1)
The correct answer is 10
= 10.009 = 10 cm ∴ The radius of path formed will be 10 cm.
38
PYQ 2022
medium
physicsID: jee-main
Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 30 \mu T. The equal current flowing in the two conductors is:
1
30 A in the same direction
2
30 A in the opposite direction
3
60 A in the opposite direction
4
300 A in the opposite direction
Official Solution
Correct Option:
(2)
The correct answer is (B) : 30 A in the opposite direction If currents are equal in both wires and field at mid-point is not zero, then the current is in opposite direction to sum up the fields at the point.
Fig.
As Bnet ≠ 0 that is the wires are carrying current in opposite direction.
in opposite direction.
39
PYQ 2022
medium
physicsID: jee-main
A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is
1
1:4
2
4:1
3
2:1
4
1:2
Official Solution
Correct Option:
(3)
Radius, The ratio of the radii of the circular paths,
So, Answer is (C):
40
PYQ 2023
medium
physicsID: jee-main
A solenoid of tums is wound uniformly in a single layer on a glass tube long and in diameter The magnetic intensity at the center of the solenoid when a current of flows through it is:
1
2
3
4
Official Solution
Correct Option:
(2)
1. The magnetic intensity inside a solenoid is given by: where is the number of turns per unit length and is the current. 2. Calculate : 3. Substituting and : Thus, the magnetic intensity is . The magnetic intensity depends on the number of turns per unit length and the current flowing through the solenoid.
41
PYQ 2023
hard
physicsID: jee-main
A charge particle of accelerated by a potential difference of enters a region of uniform magnetic field of magnitude at right angle to the direction of field The charge particle completes semicircle of radius inside magnetic field The mass of the charge particle is _______
Official Solution
Correct Option:
(1)
The correct answer is 144.
=144×10−18kg
42
PYQ 2023
easy
physicsID: jee-main
A long solenoid is formed by winding 70 turns/cm. If 2.0 A current flows, then the magnetic field produced inside the solenoid is:
1
2
3
4
Official Solution
Correct Option:
(4)
The magnetic field inside a long solenoid is given by:
where:
= (permeability of free space),
(current).
Substitute the values:
Thus, the magnetic field produced inside the solenoid is .
43
PYQ 2023
hard
physicsID: jee-main
The electric current in a circular coil of 2 turns produces a magnetic induction at its centre The coil is unwound and is rewound into a circular coil of 5 turns and the same current produces a magnetic induction at its centre The ratio of is :
1
2
3
4
Official Solution
Correct Option:
(2)
B=2RNμ0i B1=2R1N1μ0i
For N2=5
Radius of coil =R2=N2N1×R1 B2=R2N2μ0i B1B2=N1N2⋅R2R1=N1N2×N1N2;B1B2=425
44
PYQ 2023
medium
physicsID: jee-main
An electron with energy moves at right angle to the earth's magnetic field of The frequency of revolution of the electron will be (Take mass of electron )
1
2
3
4
Official Solution
Correct Option:
(3)
f=T1=2πmeB =2π×9×10−311.6×10−19×10−4=2.8×106Hz
45
PYQ 2023
easy
physicsID: jee-main
The magnetic moments associated with two closely wound circular coils and of radius and respectively are equal if : (Where and are number of turn and current of and respectively)
1
2
3
4
Official Solution
Correct Option:
(3)
Given:
Since , we have:
Substitute the values:
Simplifying the equation:
46
PYQ 2023
medium
physicsID: jee-main
A solenoid of tums is wound uniformly in a single layer on a glass tube long and in diameter The magnetic intensity at the center of the solenoid when a current of flows through it is:
1
2
3
4
Official Solution
Correct Option:
(2)
Magnetic field at centre inside the solenoid is given by B=�0?nI
So magnetic intensity at centre H=�0?B?=nI=(21200?)(2) H=1.2�103Am-1
47
PYQ 2023
medium
physicsID: jee-main
For a moving coil galvanometer, the deflection in the coil is rad when a current of is field is and the number of turns in the coil is , the area of each tum (in ) is :
1
2
3
4
Official Solution
Correct Option:
(2)
The correct answer is (B) : 1.0 τ=Kθ NiAB=Kθ A=NiBKθ=200×10×10−3×0.014×10−5×0.05
On solving A=10−4m2=1cm2
48
PYQ 2023
medium
physicsID: jee-main
The characteristics of two coil is given below. If the magnetic moment of both coil A and B are equal then choose the correct relation,
Coil A
Coil B
Radius rA=10 cm
rB=20 cm
Number of turns NA
NB
Current IA
IB
1
2NAIA = NBIB
2
NAIA = NBIB
3
NAIA = 4NBIB
4
NAIA = 2NBIB
Official Solution
Correct Option:
(3)
The correct option is (C): NAIA = 4NBIB
49
PYQ 2023
medium
physicsID: jee-main
In a moving coil galvanometer if the number of turns increases by 25%, then change in voltage sensitivity is?
1
0
2
1%
3
25%
4
50%
Official Solution
Correct Option:
(1)
The correct option is (A): 0
50
PYQ 2023
medium
physicsID: jee-main
An electron is moving along positive x direction in xy plane, magnetic field points in negative z direction, then the force due to magnetic field on electron points in the direction
1
j
2
-j
3
k
4
-k
Official Solution
Correct Option:
(2)
The power dissipated across a resistor is given by the formula:
where is the power, is the potential difference, and is the resistance. If the potential difference is doubled, the new potential difference is . The new power is: Thus, the power dissipated becomes four times when the potential difference is doubled. Hence, the correct answer is .
51
PYQ 2023
medium
physicsID: jee-main
Two long parallel wires carrying currents and in opposite directions are placed at a distance of from each other A point is at equidistant from both the wires such that the lines joining the point to the wires are perpendicular to each other The magnitude of magnetic field at is___ (Given : )
Official Solution
Correct Option:
(1)
The correct answer is 68
Magnetic fields due to both wires will be perpendicular to each other. B1=2πdμ0i1B2=2πdμ0i2 Bnet=B12+B22⇒2πdμ0i12+i22 ⇒2π×(7/2)×10−24π×10−7×82+152(d=27cm) ⇒68×10−6T
52
PYQ 2023
medium
physicsID: jee-main
A dipole having dipole moment M is placed in two magnetic fields of strength B1 and B2 respectively. The dipole oscillates 60 times in 20 seconds in the magnetic field and 60 oscillations in 30 seconds in the magnetic field. Then find the
1
2
3
4
Official Solution
Correct Option:
(4)
The time period of oscillation of a dipole in a magnetic field is given by:
where I is the moment of inertia of the dipole about its axis of rotation. For a given dipole, the moment of inertia and the dipole moment are constant. Therefore, the time period of oscillation is directly proportional to the square root of the magnetic field strength. Let the time period of oscillation in the B1 magnetic field be T1 and in the B2 magnetic field be T2. Then:
60 oscillations in 20 seconds in the B1 magnetic field gives:
60 oscillations in 30 seconds in the B2 magnetic field gives:
Substituting these values in the above equation, we get:
Squaring both sides, we get:
Therefore, . Answer. D
53
PYQ 2023
easy
physicsID: jee-main
Two identical current carrying coils with same centre are placed with their planes perpendicular to each other. If i = and radius of coil R = 1 m, then magnetic field at centre C is equal to:
1
2
3
4
Official Solution
Correct Option:
(1)
So, the correct answer is (A): μ0
54
PYQ 2023
hard
physicsID: jee-main
Two long straight wires and carrying equal current each were kept parallel to each other at distance Magnitude of magnetic force experienced by length of wire is If distance between wires is halved and currents on them are doubled, force on length of wire will be:
1
2
3
4
Official Solution
Correct Option:
(2)
Force per unit length between two parallel straight wires:
The force per unit length between two parallel wires carrying currents and is given by the formula:
Given that the currents in two cases are A and A, and the separation distance is cm in the first case and cm in the second case, the force ratio can be calculated as:
Simplifying the expression:
Therefore, we get:
55
PYQ 2023
medium
physicsID: jee-main
Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion If the ratio of radii of their circular paths is and their respective masses ratio is Then, the ratio of their charges will be :
1
2
3
4
Official Solution
Correct Option:
(2)
The correct option is (B)
56
PYQ 2023
medium
physicsID: jee-main
A metallic cube of side 15 cm moving along y-axis at a uniform velocity of 2 ms–1. In a region of uniform magnetic field of magnitude 0.5T directed along z-axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be ___ mV.
Official Solution
Correct Option:
(1)
Given:
Side of the cube ( ) =
Velocity ( ) =
Magnetic field ( ) =
Step 1: Formula for Induced EMF
The induced electromotive force (EMF) or potential difference ( ) across the faces of the cube due to its motion in the magnetic field is given by:
where:
: Magnetic field strength
: Length of the side of the cube perpendicular to both velocity and magnetic field
: Velocity of the cube
Step 2: Substitute the Given Values
Substitute , , and into the formula:
Simplify:
Convert to millivolts:
Final Answer:
The potential difference developed between the faces is .
57
PYQ 2023
medium
physicsID: jee-main
A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is
1
zero in the region r < a and inversely proportional to r in the region r > a
2
inversely proportional to r in the region r < a and uniform throughout in the region r > a
3
directly proportional to r in the region r < a and inversely proportional to r in the region r > a
4
uniform in the region r < a and inversely proportional to distance r from the axis, in the region r > a
Official Solution
Correct Option:
(3)
It is a case of solid infinite current carrying wire. Using ampere circuital law, CASE I: if r ≤ R 𝐵 = 𝜇_0𝑖/2𝜋𝑅^2 𝑟 CASE II: r>R 𝐵 = 𝜇0𝑖 /2𝜋r
58
PYQ 2024
easy
physicsID: jee-main
The magnetic field existing in a region is given by A square loop of edge carrying current is placed in the - plane with its edges parallel to the - and -axes, as shown in the figure. The magnitude of the net magnetic force experienced by the loop is ______ .
Official Solution
Correct Option:
(1)
Magnetic Force on a Current-Carrying Wire in a Magnetic Field: The magnetic force on a segment of current-carrying wire in a magnetic field is given by: where: is the current, is the length of the segment, is the magnetic field, is the angle between the magnetic field and the current direction.
In this case, the loop lies in the plane with its edges parallel to the and axes. Since varies with , the magnetic force on each side of the loop depends on its position in the plane.
Calculate the Force on Each Side of the Loop: For the left side at :
Force on the left side:
For the right side at :
Force on the right side:
Net Force on the Loop: The forces on the top and bottom sides (parallel to the -axis) will cancel each other out due to symmetry, as the magnetic field along these sides is the same. Therefore, the net force is due to the difference in the forces on the left and right sides:
Convert to mN:
Conclusion: The magnitude of the net magnetic force experienced by the loop is .
59
PYQ 2025
medium
physicsID: jee-main
Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is T. The value of , which represents the distance of point P from wire X, is cm. (Take permeability of free space as SI units.)
Official Solution
Correct Option:
(1)
To find the distance of point P from wire X, we need to consider the magnetic fields produced by both wires at point P. The magnetic field due to a straight current-carrying wire at a distance is given by:
For wire Y, at distance 4 cm:
For wire X, at distance :
Since the currents are in opposite directions, their magnetic fields at point P are in opposite directions. Hence, the resultant magnetic field is:
Substituting values:
Consider :
60
PYQ 2026
medium
physicsID: jee-main
A small cube of side mm is placed at the centre of a circular loop of radius cm carrying a current of A. The magnetic energy stored inside the cube is J. The value of is ————.
1
6.28
2
3
628
4
Official Solution
Correct Option:
(1)
To find the magnetic energy stored in a volume, we use the magnetic energy density formula. The energy density (energy per unit volume) in a magnetic field is given by: $ B R I I = 2 R = 10 = 0.1 \mu_0 = 4\pi \times 10^{-7} u_B U a = 1 = 10^{-3} V = a^3 \pi = 3.14 \alpha \times 10^{-14} \alpha = 6.28$.
61
PYQ 2026
easy
physicsID: jee-main
A current carrying solenoid is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy:
1
2
3
4
Official Solution
Correct Option:
(4)
Step 1: Understanding the Question:
A charged particle is released from rest inside a vertical current-carrying solenoid. We need to determine its acceleration. The key is to analyze the forces acting on the particle. Step 2: Key Formula or Approach:
The forces on the particle are:
1. Gravitational force: (acting vertically downwards).
2. Magnetic force (Lorentz force): . The net force determines the acceleration: . Step 3: Detailed Explanation:
1. Magnetic Field Direction: The magnetic field ( ) inside a long solenoid is uniform and directed along its axis. Since the solenoid is placed vertically, the magnetic field is also vertical. 2. Initial Motion: The particle is released from rest, so its initial velocity . At this instant, the magnetic force is . The only force is gravity, so the particle begins to accelerate downwards due to gravity. 3. Subsequent Motion: As the particle starts moving, its velocity vector will be directed vertically downwards, along the axis of the solenoid. 4. Calculating Magnetic Force: The velocity vector and the magnetic field vector are both along the vertical axis. This means they are parallel to each other. The angle between and is 0° or 180°.
The magnitude of the magnetic force is given by .
Since and , the magnetic force on the particle is always zero as long as it moves along the axis. 5. Net Force and Acceleration: Since the magnetic force is always zero, the only force acting on the particle is the gravitational force, .
From Newton's second law: Step 4: Final Answer:
The acceleration of the charged particle is equal to the acceleration due to gravity, g. This corresponds to option (D).
62
PYQ 2026
hard
physicsID: jee-main
Consider two arrangements of wires. Find the ratio of magnetic field at the centre of the semi–circular part.
1
(A)
2
(B)
3
(C)
4
(D)
Official Solution
Correct Option:
(1)
Magnetic field at the centre is calculated using the Biot–Savart law. For a semicircular wire of radius , . For the first arrangement, the fields from the straight sections and the curve add up. In the second, some components subtract.
Evaluating the vector sum for both geometries and taking the ratio leads to the expression .
63
PYQ 2026
medium
physicsID: jee-main
The equilateral triangular frame has current 2A. The side of frame is cm. Magnetic field at center C is:
1
2
3
4
Official Solution
Correct Option:
(1)
Step 1: Understanding the Question:
We need to find the total magnetic field at the center (centroid) of an equilateral triangular loop carrying a current. The total field is the vector sum of the fields produced by the three straight wire segments of the triangle. Step 2: Key Formula or Approach:
The magnetic field at a perpendicular distance from a finite straight wire carrying current is given by:
where and are the angles subtended by the ends of the wire at the point. Step 3: Detailed Explanation: Part A: Geometric Parameters
- Side length of the equilateral triangle, cm = m.
- The center C is the centroid. The perpendicular distance from the centroid to any side is given by .
- For an equilateral triangle, the angles subtended by the ends of any side at the centroid are and . Part B: Magnetic Field from One Side
Using the formula for a finite wire with A: Part C: Total Magnetic Field
By the right-hand rule, the magnetic field from each of the three sides points in the same direction at the center (e.g., into the page). Therefore, the net magnetic field is the sum of the magnitudes of the fields from the three sides.
To express this in microteslas ( ), we multiply by . Step 4: Final Answer:
The magnetic field at the center C is .
64
PYQ 2026
medium
physicsID: jee-main
Three very long parallel wires carrying current as shown. Find the force acting at 15 cm length of middle wire :
1
1 N
2
6 N
3
7 N
4
5 N
Official Solution
Correct Option:
(2)
Step 1: Understanding the Concept: Parallel wires carrying currents in the same direction attract each other. Parallel wires carrying currents in opposite directions repel each other. Step 2: Key Formula or Approach: The magnetic force per unit length between two wires is:
Force on length : . Step 3: Detailed Explanation: Let the wires be A (3A), B (1A), and C (2A). We need the force on wire B. Distance . Distance . Length . 1. Force on B due to A ( ): Currents are opposite (3A up, 1A down), so it is repulsive. Wire B is pushed to the right by wire A.
2. Force on B due to C ( ): Currents are in the same direction (1A down, 2A down), so it is attractive. Wire B is pulled to the right by wire C.
Since both forces act to the right, the net force is:
Step 4: Final Answer: The net force acting on the 15 cm length of the middle wire is 6 N.
65
PYQ 2026
medium
physicsID: jee-main
A circular coil of radius R carries current such that magnetic field at its centre is 16 T. Find the magnetic field on the axis at a distance of from the centre of coil.
1
2 T
2
4 T
3
3 T
4
5 T
5
None of these
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: The magnetic field of a current loop decreases as we move along its central axis away from the center. Step 2: Key Formula or Approach: Field at center: . Field on axis at distance : . Ratio: . Step 3: Detailed Explanation: Given .
Given T}\) :
Step 4: Final Answer: The magnetic field at the given axial point is 2 T.
66
PYQ 2026
medium
physicsID: jee-main
A bar magnet is kept such that it is making an angle of 30 with the magnetic field. The torque acting on the magnet is 0.016 N-m. Find the amount of work done by external agent in rotating the magnet from most stable position to most unstable position.
1
0.064 J
2
0.020 J
3
0.034 J
4
0.055 J
Official Solution
Correct Option:
(1)
Step 1: Understanding the Question:
A bar magnet experiences a torque when placed in a uniform magnetic field at a given angle. We need to use this information to find the magnetic moment or the product (or MB). Then, we must calculate the work done to rotate the magnet from its most stable equilibrium to its most unstable equilibrium. Step 2: Key Formula or Approach:
1. Torque on a magnetic dipole: , where is the magnetic moment.
2. Potential energy of a magnetic dipole: .
3. Most stable position is at (Minimum PE).
4. Most unstable position is at (Maximum PE).
5. Work done by external agent: . Step 3: Detailed Explanation:
Given:
Initial angle,
Torque, Using the torque formula to find :
Now, calculate the work done to rotate the magnet from (most stable) to (most unstable).
Since and :
Substitute the value of MB:
Step 4: Final Answer:
The amount of work done by the external agent is 0.064 J.
67
PYQ 2026
easy
physicsID: jee-main
An ideal solenoid is kept with its axis vertical. Current is flowing in the solenoid. A charge is thrown downward inside the solenoid. The acceleration of the charged particle is then
1
2
3
4
Official Solution
Correct Option:
(2)
Concept: An ideal solenoid produces a uniform magnetic field inside it, directed along its axis. A magnetic field exerts a force on a moving charged particle only if the velocity of the particle has a component perpendicular to the magnetic field. The magnetic force on a charge is given by: Step 1: Direction of magnetic field and velocity Inside an ideal solenoid, the magnetic field is along the axis of the solenoid. Since the solenoid’s axis is vertical, is vertical. The charged particle is thrown downward, so its velocity is also vertical. Step 2: Magnetic force on the charged particle Because is parallel to , Hence, the magnetic force acting on the charge is zero. Step 3: Net force and acceleration Since no magnetic force acts on the particle, the only force acting on it is gravity: Therefore, the acceleration of the particle is: Conclusion: The acceleration of the charged particle remains equal to gravitational acceleration.
68
PYQ 2026
medium
physicsID: jee-main
For a circular coil of radius , magnetic field at the center is .
What will be the magnetic field on the axis at a distance from the center?
1
2
3
4
Official Solution
Correct Option:
(4)
Concept:
The magnetic field on the axis of a circular current-carrying coil at a distance from its center is given by:
where:
is the magnetic field at the center of the coil, is the radius of the coil, is the axial distance from the center.
Step 1: Substitute the given values. Step 2: Calculate the magnetic field at the given point. Conclusion:
Hence, the correct answer is (4).
About Moving Charges And Magnetism - JEE-MAIN
Moving Charges And Magnetism is a vital chapter for JEE-MAIN aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
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