Moving Charges And Magnetism

The Lorentz force is given by: First, calculate the cross product : Now, adding : Multiplying by : Magnitude:

We are tasked with finding the initial moment of force (torque) acting on a rectangular coil placed in a uniform magnetic field. The given parameters are: Number of turns, ,
Area of the coil, ,
Current through the coil, ,
Magnetic field strength, ,
Angle between the plane of the coil and the magnetic field, . Step 1: Recall the formula for torque on a current-carrying coil The torque ( ) acting on a current-carrying coil in a magnetic field is given by: where:
is the number of turns,
is the current,
is the area of the coil,
is the magnetic field strength,
is the angle between the magnetic field and the normal to the plane of the coil.
Step 2: Determine the angle The angle is the angle between the plane of the coil and the magnetic field. The angle is the angle between the normal to the plane of the coil and the magnetic field. These two angles are complementary, so: Step 3: Substitute values into the torque formula Substitute the given values into the torque formula: Simplify: Thus, the initial moment of force acting on the coil is 1 Nm.

The relation between electric field amplitude and magnetic field amplitude in an electromagnetic wave is: where m/s and T.

Diamagnetic materials are those which are repelled by a magnetic field. Superconductors exhibit perfect diamagnetism, meaning they expel all magnetic fields when cooled below a certain temperature, which is why they are considered one of the most exotic diamagnetic materials.

Step 1:
The magnetic field at the center of the loop is given by: where is the current and is the radius of the loop.
Step 2:
The magnetic field at a distance on the axis of the loop is given by:
Step 3:
The ratio of the magnetic fields is:

The mutual inductance between two circular coils is given by: where and are the radii of the coils, and implies that the first coil is much smaller than the second coil.

Step 1:
The magnetic force on a charged particle moving in a circular path is given by: where: - is the mass of the particle,
- is its velocity,
- is the radius of the circular path.
The force experienced by the particle is also equal to the magnetic force: where:
- is the charge,
- is the magnetic field strength.
Equating the two expressions for force gives:
Step 2:
Solving for the linear momentum , we get: Since the velocity is proportional to the radius and the charge, the ratio of linear momenta of the two particles is proportional to the product of the charge and mass ratio.
Step 3:
Let the charge and mass ratio of the two particles be and . The ratio of the linear momenta is given by:

The magnetic field (B) at a distance r from a long straight wire carrying current I is given by Ampere's law: .
The value of the constant is T·m/A.
Let's call the wire with 8 A current wire 1, and the wire with 10 A current wire 2.
A, A. The currents are in opposite directions.
The point P is between the wires. The distance from wire 1 is .
The total separation is 9 cm, so the distance from wire 2 is .
At a point between two wires carrying opposite currents, the magnetic fields from both wires point in the same direction (by the right-hand grip rule). Therefore, the net magnetic field is the sum of the individual fields.
Magnetic field from wire 1: T.
Magnetic field from wire 2: T.
The net magnetic field is the sum of and .
T.

The magnetic moment ( ) of a circular coil is given by the formula , where n is the number of turns, I is the current, and A is the area of the coil.
The area of a circular coil with radius r is .
So, the magnetic moment is .
Let's write down the information given for coil A and coil B.
For coil A: , , , .
For coil B: , , , .
We are given that the currents are equal: .
We are given the ratio of their magnetic moments: .
Now, let's write the ratio of the magnetic moments using the formula.
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Since and is a constant, they cancel out.
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We are given this ratio is 1/2. Substitute the known values.
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Now, we need to solve for the ratio of the radii, .
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Taking the square root of both sides:
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The ratio of the radii of coils A and B is .

The potential energy (U) of a magnetic dipole with moment M in a uniform magnetic field B is given by , where is the angle between the magnetic moment and the magnetic field.
The most stable position occurs when the potential energy is minimum. This happens when is maximum, i.e., , which corresponds to . In this position, the magnetic moment is aligned with the field.
The potential energy in the most stable position is .
The most unstable position occurs when the potential energy is maximum. This happens when is minimum, i.e., , which corresponds to . In this position, the magnetic moment is anti-aligned with the field.
The potential energy in the most unstable position is .
The work done in moving the magnet from the stable to the unstable position is the change in its potential energy.
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We are given the values:
Magnetic moment, Am .
Magnetic field, T.
Substitute these values to find the work done.
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J.

Step 1: Relation between N and r:
Let the length of the wire be . For coil A with turns of radius : . For coil B with turns of radius : .
Step 2: Magnetic Field Formula:
Magnetic field at the center of a coil with turns: . Since is same, . Substituting , we get .
Step 3: Calculate Ratio:

Step 1: Calculate Loop Dimensions:
Total length cm = 0.9 m. Equilateral triangle side m. Area of loop .
Step 2: Magnetic Moment ( ):
(Here ). .
Step 3: Calculate Torque ( ):
Formula: . Important: is the angle between the area vector (normal to loop) and the magnetic field. Given: Plane of loop makes with field. So, . The 8 in numerator cancels in denominator.
Step 4: Final Answer:
The torque is Nm.

Step 1: Determine the length of the side of the equilateral triangle.
The total length of the wire is cm. The triangle has 3 equal sides, so the length of each side ( ) is: Convert to SI units (meters): .

Step 2: State the formula for the magnetic force on a current-carrying wire.
The magnitude of the magnetic force on a segment of wire of length carrying a current in a uniform magnetic field is given by: where is the angle between the current direction (along the wire) and the magnetic field .

Step 3: Determine the angle for the side AC.
The magnetic field is applied parallel to the side BC. The angle between side AC and side BC in an equilateral triangle is (or rad). The current flows along the sides, and the magnetic field is parallel to BC. The angle between the current direction on side AC and the magnetic field (parallel to BC) is .

Step 4: Substitute the given values and calculate the force.
Given values: A, m, T, and .

Step 1: State the formula for the radius of a charged particle's path in a magnetic field.
When a charged particle enters a uniform magnetic field perpendicularly ( ), the magnetic force provides the centripetal force, and the particle moves in a circular path. The radius of the path is given by: where is the mass, is the velocity, is the charge, and is the magnetic field magnitude.

Step 2: Solve the formula for the magnetic field .
Rearrange the formula to isolate :

Step 3: Convert the given values to standard units (SI).
Mass of proton: kg. Velocity: m/s. Charge of proton: C. Radius of path: .

Step 4: Substitute the numerical values and calculate B.
Group the numerical and exponential parts: Simplify the numerical part: Simplify the exponential part:

Step 5: Convert the result to mT.
Since :

Step 1: Analyze the Circuit:
The figure shows a square loop of uniform wire. Current enters at one corner and leaves at another. Let the side of the square be . The total current is . Let the input terminal be A and the output terminal be B. The current splits into two paths: 1. Short path (one side, length , Resistance ). 2. Long path (three sides, length , Resistance ).
Step 2: Current Division:
Since the wire is uniform, resistance is proportional to length. Current in short path ( ) and long path ( ) divides inversely to resistance. . .
Step 3: Magnetic Field Calculation:
The magnetic field at the center of a square side of length carrying current is . Field due to short path (1 side): . (Direction: Out of page, say). Field due to long path (3 sides): . (Direction: Into page).
Step 4: Net Field:
The magnitudes are equal: . The directions produced by the currents in the two branches at the center are opposite (one clockwise, one counter-clockwise relative to center). Thus, the net magnetic field is . Note:
This result holds true for any regular polygon loop of uniform wire where current enters and leaves at any two vertices.

Step 1: Formulae:
Angle of dip . Vertical component . Horizontal component . Total magnetic field .
Step 2: Express B in terms of components:
From the vertical component equation: .
Step 3: Substitute :
. (Checking horizontal option: . This is not among the options in that form).

Step 1: Identify Flux Angles:
The magnetic flux is , where is the angle between the normal to the coil and the magnetic field. Initial state: Plane makes with field. So normal makes . Final state: Plane is parallel to field. So normal makes .
Step 2: Calculate Flux Change:
, T, . Wb. . Change in flux Wb.
Step 3: Calculate Induced EMF and Current:
Time interval s. Induced EMF V. Induced Current A.

Step 1: AC Circuit Formula:
In a series RL circuit connected to an AC source, the voltages across the resistor ( ) and the inductor ( ) are not in phase. leads by . The total supply voltage is the phasor sum:
Step 2: Calculation:
Given V and V.

Step 1: Formula for Radiation Pressure Force:
When radiation is absorbed (non-reflecting surface), the momentum transfer per unit time is . The force exerted is: where is the speed of light ( m/s).
Step 2: Calculation:
Power W.

The magnetic force on a charged particle is given by the Lorentz force law: .
Here, the particle is an alpha particle, so its charge is positive ( ).
The velocity vector is directed towards the east.
The magnetic field vector is directed vertically up.
The velocity vector and the magnetic field vector are perpendicular to each other.
According to the formula for the cross product, the force vector is perpendicular to both and .
Using the right-hand rule for a positive charge: Point fingers in the direction of velocity (East). Curl fingers in the direction of the magnetic field (Up). The thumb points in the direction of the force. This direction is South.
The force is initially directed towards the South. The velocity is East. Since the force is perpendicular to the velocity, it acts as a centripetal force, causing the particle to move in a circular path.
The force vector lies in the East-South direction, which is horizontal. Since the force is always in the horizontal plane (perpendicular to the vertical B-field), the particle's motion will be confined to a horizontal circular path.
The magnetic force does no work on the charged particle because the force is always perpendicular to the direction of motion (velocity).
Since no work is done, the kinetic energy of the particle remains constant. Therefore, the speed of the particle remains the same.
Combining these findings, the alpha particle will move in a horizontal circular path with the same speed.

The voltage sensitivity ( ) of a moving coil galvanometer is defined as the deflection per unit voltage.
, where is the deflection and is the voltage.
The deflection is given by , where N is the number of turns, A is the area, B is the magnetic field, k is the torsional constant, and I is the current.
Voltage , where R is the resistance of the coil.
Substituting these into the sensitivity formula:
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We are given that all other quantities (B and k) remain the same. So, we can write the proportionality:
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We are given the ratios for galvanometer A and B:
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We can set up a ratio of the sensitivities using the proportionality:
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Now, substitute the given ratios:
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The term cancels from both sides:
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This gives the relationship between the number of turns: .
We are given that the number of turns for galvanometer A is .
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Step 1: Understanding the Concept:
A current-carrying loop placed in a magnetic field experiences a torque. The magnitude of this torque depends on the magnetic moment of the loop, the strength of the magnetic field, and the orientation of the loop with respect to the field.
Step 2: Key Formula or Approach:
The torque ( ) on a coil is given by the vector product of its magnetic dipole moment ( ) and the magnetic field ( ):
The magnitude of the torque is:
where:
- is the magnitude of the magnetic moment.
- is the number of turns.
- is the current.
- is the area of the coil.
- is the angle between the magnetic moment vector and the magnetic field vector .
Step 3: Detailed Explanation:
First, let's list the given values in SI units:
- Side of the square coil, .
- Area of the coil, .
- Number of turns, .
- Magnetic field strength, .
- Current, .
Next, determine the angle . The problem states "the plane of the coil is in the direction of magnetic field". This means the magnetic field lines are parallel to the surface of the coil. The magnetic moment vector is, by definition, perpendicular to the plane of the coil. Therefore, the angle between and is .
Now, calculate the magnetic moment :
Finally, calculate the torque :
Since , the torque is at its maximum value.
Step 4: Final Answer:
The torque acting on the coil is Nm. Therefore, option (A) is correct.

Step 1: Understanding the Concept:
This problem involves the formulas for the magnetic field produced by a circular current loop, both at its center and at a point along its axis. We need to establish a relationship between the two.
Step 2: Key Formula or Approach:
Let be the radius of the coil.
- The magnetic field at the center of the coil is: .
- The magnetic field at a distance from the center on its axis is: .
Step 3: Detailed Explanation:
First, let's relate the given distances to the radius .
- Diameter of the coil = .
- Radius of the coil, .
- The distance from the center, .
We are given that the field at this distance is . Let's write the expression for :
Now, simplify the denominator:
So, the expression for B becomes:
We need to find the magnetic field at the center, :
To relate to , we can express from equation (1):
Now substitute this into equation (2):
Step 4: Final Answer:
The magnetic field at the centre of the coil is 27B. Therefore, option (B) is correct.

Step 1: Understanding the Concept:

We need to find the relative permeability , which is the ratio of the permeability of the substance to that of free space .
Step 2: Key Formula or Approach:

Relative permeability , where is the magnetic susceptibility. Also .
Step 3: Detailed Explanation:

Given . Convert to fraction: Thus, the ratio .
Step 4: Final Answer:

The ratio is 8:5.

Step 1: Understanding the Concept:

The magnetic field inside a long solenoid is uniform and parallel to the axis. The point 5 cm from the axis is inside the solenoid (since radius is 10 cm). The total magnetic field is due to both layers of windings.
Step 2: Key Formula or Approach:

Magnetic field inside a solenoid: . Where is turns per unit length. For multiple layers, is the total number of turns or simply add the fields if layers are in series. Here, we use total turns density.
Step 3: Detailed Explanation:

Given: Length . Total turns (two layers of 100 each). Current . Permeability . Calculate : Calculate : The question asks for the value in . Value .
Step 4: Final Answer:

The magnetic field is T.

Step 1: Understanding the Concept:

A charged particle moving perpendicular to a magnetic field executes circular motion. The time period of revolution depends on the charge-to-mass ratio but is independent of speed.
Step 2: Key Formula or Approach:

Time period of revolution: Ratio:
Step 3: Detailed Explanation:

For Proton ( ): Mass , Charge . For Alpha particle ( ): Mass (2 protons + 2 neutrons), Charge . Substitute into ratio:
Step 4: Final Answer:

The ratio is 1:2.

• Magnetic field inside toroid: .
• m, , A, T·m/A.
• T.
• Hence correct answer: in T units}.