Electromagnetic Induction

Let the current in the bigger loop be i₂ and the smaller loop be i₁

• ϕ₁ is the flux due to the smaller loop at the bigger loop
• ϕ₂ be flux due to a bigger loop at the smaller loop

The field due to the current loop 1 at an axial point -

Flux linked with the smaller loop 2 due to B₁ is

The coefficient of mutual inductance between the loops is

Flux linked with bigger loop 1 is

Substituting the given values, we get

Option B is the correct answer, ϕ1 = 9.1 x 10^-11 Weber

Total length L will remain constant

(N = total\, turns)

and length of winding ( )= (d = diameter of wire)

Self inductance =

Self inductance (L) ∝

As

So, as a is increased to 3a the Self inductance will be 3 times.

Hence, the correct answer is option (B): Increases by a factor of 3

To solve this problem, we need to address two parts: the time elapsed before the current reaches half of its steady-state value and the energy stored in the magnetic field after a certain time.

1. Steady-state current calculation:
   • The steady-state current, , can be found using Ohm's Law: .
   • Given that the voltage and resistance , we have: .
2. Time for current to reach half of the steady-state value:
   • The current at time is given by the formula: ,
   • where is the time constant.
     For and , .
   • For the current to be half its steady-state value: ,
   • This simplifies to: .
   • Taking the natural log on both sides, we get: .
   • Substituting the values, we get: .
3. Energy stored in the magnetic field after 15 ms:
   • The energy stored in the magnetic field, , is given by: .
   • Let's first find :
     For , ,
   • thus .
   • Now, substitute into the formula for energy: ,
   • which is approximately .
4. Conclusion:
   • Therefore, the time elapsed before the current acquires half of its steady-state value is approximately 7 ms, and the energy stored after 15 ms is 1 mJ.

Thus, the correct answer is: t = 7 ms; U = 1 mJ.

Emf = Bωl²
= × 0.2 × 10^–4 × 5 × 1² V
= 0.5 × 10^–4 V
= 50 μV

The correct answer is (B): 50 μV

Self-inductances are in series but their mutual inductances are linked oppositely so equivalent self-inductance
L = L₁ + L₂ – M – M
L = L₁ + L₂ – 2M

The correct option is (D):

The correct answer is (C) : [ML²T^–2A^–2]

∵ U = Mi²

⇒ [M] = =
= [ML²T^–2A^–2]

In a meter bridge setup, the balance condition for measuring resistance is given by the equation:

S/R = l/(100-l)

where S is the unknown resistance, R is the known resistance (5.6 kΩ), l is the length of the wire from one end to the null point (30 cm), and (100-l) is the remaining length (70 cm).

Now, let's calculate the unknown resistance S:

S = R × l/(100-l)

Substitute the values:

S = 5600 × 30/70

S = 2400 Ω

Upon calculating, the value of S is 2400 Ω, which fits within the given range of [2400, 2400]. Thus, the calculated unknown resistance is confirmed to be accurate.

To solve this problem, we need to understand how changes in the coil's dimensions affect the electrical power dissipated due to the current induced. This involves Faraday's law of electromagnetic induction and the resistance formula.

1. **Faraday's Law:** The induced electromotive force (EMF) in the coil due to a change in magnetic flux is given by:

where is the EMF, is the number of turns, and is the rate of change of magnetic flux.

2. **Resistance of the Coil:** The resistance of the coil is given by:

where is the resistivity, is the length of the wire, and is the cross-sectional area of the wire. The cross-sectional area for a wire of radius is . With the radius doubled, becomes .

3. **Effect on Resistance:** Given that the radius is doubled, the new resistance becomes:

4. **Induced Current**: The induced current can be expressed as:

or
As is halved,

5. **Electrical Power Dissipated**: Power dissipated is given by . For the new configuration:

Substituting the expressions for and :

This indicates the power is doubled.

Conclusion: The correct answer is Doubled, confirming that the electrical power dissipated in the coil, under the given changes, is doubled.

The correct answer is (A) :

To find the magnetic moment of a wire bent to form a circle, we first need the formula for the magnetic moment of a coil: , where is the number of turns, is the current, and is the area of the circle.
1. The wire length is given as 314 cm. When bent into a circle, the circumference of the circle is the length of the wire, so .
2. Solving for the radius gives: cm.
3. Convert radius into meters: cm = m.
4. Calculate the area of the circle: m².
5. Since it's a single loop, .
6. The current is given as 14 A.
7. Substitute the values to find the magnetic moment: .
8. Given range is from 11 to 11, hence is approximately equal to 11 A-m² which is within the accepted range.

ϕ = μ_r​μ₀​ ​I × A

ϕ = 4π × 10−⁶ x 4π × 10⁻⁷ x 0.4 x 2 x 10 ^-4
μ_r​=125

So, the correct answer is (A): 125

The correct answer is (B) :

The relationship between the charges and radii is given by:

Simplifying, we find:

Using the charge conservation equation:

Substitute :

Substitute :

The surface charge densities are related by:

Electromagnetic Force and Voltage:

The force on a current-carrying conductor in a magnetic field is given by:

Equating with gravitational force :

Substitute :

Solve for :

Substitute the given values , , , , and :

Final Answer:

Voltage

Given:

Correct Answer: (A)

The formulae involve relationships for current and voltage sensitivities. Equation (1) gives the relationship between the magnetic field and the area of the coil . From Equation (2), we can deduce that voltage sensitivity is directly proportional to current sensitivity. This shows a clear relation where remains constant.

Given:

• Inductance ( ) = 2 H
• Resistance ( ) = 4
• Applied voltage ( ) = 10 V

Step 1: Analyze the Circuit at Steady State

At steady state, the inductor behaves as a short circuit, meaning it has zero resistance. The circuit becomes a simple resistor connected to the voltage source.

Step 2: Calculate the Current at Steady State

Using Ohm’s law, the steady-state current ( ) through the resistor is given by:

Substitute and :

Step 3: Calculate the Energy Stored in the Magnetic Field

The energy ( ) stored in the magnetic field of the inductor is given by:

Substitute and :

Calculate :

Substitute back into the equation:

Convert to scientific notation:

Final Answer:

The energy stored in the magnetic field is .

The magnetic field is initially given by the equation:

Step 1: , which is the original magnetic field expression.

Step 2: For the new radius , the new field is given by:

Step 3: Substitute into the magnetic field formula: .

Step 4: Now, calculate the ratio , indicating that the new magnetic field is of the original field.

Step 5: With this ratio, we conclude that the new magnetic field is .

- Magnetic field contributions due to segments and are outward at point .

Total magnetic field:

Electromagnetic Force and Current:

The force on a current-carrying conductor in a magnetic field is given by:

Equating with the gravitational force , we get:

Solving for :

Substitute the given values:

Voltage Across the Loop:

The resistance of the loop is given as . Using Ohm's Law:

Substitute and :

Final Answer:

Step 1: Write the formula for work done:
The work done ( ) in pulling the loop is related to the induced emf and the resistance of the loop: where: (magnetic field strength), (velocity of pulling the loop), (length of one side of the loop, calculated as ), \item (resistance of the loop).
Step 2: Substitute the values into the formula:
Step 3: Convert to millijoules:

Electromagnetic Induction Problem

Step 1: Faraday's Law of Electromagnetic Induction

Faraday's law states that an emf is induced in a coil when the magnetic flux through the coil changes with time.

Step 2: Magnetic Flux

Magnetic flux ( ) is given by:

where is the magnetic field, is the area vector of the coil, and is the angle between the magnetic field and the area vector.

Step 3: Analyze Options

• A. Moving with uniform speed: If the coil moves with uniform speed in a uniform magnetic field, the flux remains constant (assuming the orientation of the coil relative to the field remains unchanged). Therefore, no emf is induced.
• B. Moving with non-uniform speed: Similar to case A, if the orientation doesn’t change with respect to the field, a non-uniform speed doesn’t change the flux. Thus no emf is induced.
• C. Rotating the coil: When the coil rotates in a uniform magnetic field, the angle between the magnetic field and the area vector changes. This changes the magnetic flux, inducing an emf.
• D. Changing the area: Changing the area of the coil directly changes the magnetic flux, inducing an emf.

Conclusion:

An induced emf can be produced by rotating the coil (C) and by changing the area of the coil (D) (Option 2).

Induced emf can be induced in a coil by changing magnetic flux. And 𝜙 = 𝐵⃗ . 𝑑𝐴⃗⃗⃗⃗⃗ By rotating coil, angle between coil and magnetic field changes and hence flux changes. By changing area, magnetic flux changes.

To find the number of turns in the coil given the information about the change in the magnetic field, we can use Faraday's Law of Electromagnetic Induction, which states that the induced emf ( ) in a coil is proportional to the rate of change of magnetic flux through the coil. The formula is:

where:

• is the induced emf,
• is the number of turns in the coil,
• is the change in magnetic flux,
• is the time duration of the change.

The magnetic flux through the coil is given by:

where:

• is the magnetic field,
• is the area of the coil.

The area of the coil can be calculated from its diameter. Given the diameter , the radius . Thus, the area is:

The change in magnetic field is:

The change in magnetic flux is:

The time duration . Since the induced emf , we can write:

Solving for :

Calculating the value:

Thus, the number of turns in the coil is 70.

Step 1. Calculate the Effective Vertical Component of the Magnetic Field:

Given:

The vertical component of the magnetic field , considering the angle of dip , is:

Step 2. Convert Angular Velocity from rpm to rad/s:

Angular velocity in rad/s is given by:

Step 3. Determine the Radius of Rotation:

The length of each blade is . Therefore, the effective radius of rotation is:

Step 4. Calculate the Induced emf:

The emf induced across the tips of the blades (assuming the emf induced across two opposite ends) is given by:

Substituting the values:

Simplifying further:

Step 5. Conclude the Value of :

Comparing with , we find:

To find the power consumed by the load resistance connected to the secondary side of the transformer, we need to follow these steps:

1. Calculate the voltage across the secondary winding using the turns ratio. The turns ratio of the transformer is given as 10:1, meaning the primary winding has 10 times more turns than the secondary winding.
2. The voltage across the secondary ( ) can be calculated using the formula for voltage transformation in transformers:

1. where is the primary voltage.

1. Using Ohm's Law, calculate the current flowing through the load resistance:

1. where .

1. Calculate the power consumed by the load:

Therefore, the power consumed in the load resistance is 11.5 W.

The correct answer is 11.5 W.

To calculate the average electromotive force (emf) induced in the loop, we can use Faraday's Law of electromagnetic induction, which states that the induced emf in a closed loop is equal to the negative rate of change of magnetic flux through the loop. Mathematically, this is given by:

where is the change in magnetic flux and is the change in time.

1. Calculate the initial magnetic flux ( ):
The magnetic flux through the loop is given by:

where (magnetic field strength), (area of the loop), and (angle with the magnetic field).

Plugging in the values, we get:

2. Calculate the final magnetic flux ( ):
When the loop is removed from the field, , hence .

3. Calculate the change in magnetic flux ( ):

4. Calculate the average emf:
Given that , the average emf is:

Thus, the average emf induced in the loop during this time is . Therefore, the correct answer is .

The mutual inductance between two coils is given as . The current in the first coil changes according to , where and . We need to determine the maximum emf ( ) induced in the second coil and find the value of when .

According to Faraday's Law of electromagnetic induction, the emf induced in the second coil is given by:

Where is the rate of change of current in the first coil.

Now, let's calculate :

The maximum value of is 1, so the maximum rate of change of current is .

Substituting into the emf formula, we get:

Ignoring the negative sign, . Given , we equate and solve:

The computed value of is 2, which falls within the expected range [2, 2].

Thus, the value of is 2.

The power dissipated in the loop can be calculated using the formula:

Where is the induced current and is the resistance.

The magnetic flux change through the loop is given by:

Using Faraday’s Law, the induced emf in the loop is:

Now, calculate the induced current :

Substituting the values, we find the power dissipated:

Step 1: Magnetic Field (MF) due to the larger square loop at the center

The magnetic field at the center of a square loop is given by:

Simplifying the expression, we get:

Step 2: Magnetic Flux ( ) in the smaller coil

The magnetic flux in the smaller coil is the product of the magnetic field and the area of the coil , which gives:

Simplifying, we have:

Step 3: Mutual Inductance (M)

The mutual inductance is given by the ratio of flux to the current :

Substituting the values, we get:

Therefore, the value of is:

The magnetic flux ( ) through loop B due to current in loop A is given by:

The mutual inductance is:

where is the radius of loop A, is the distance between the loops, and is the permeability of free space.