Electromagnetic Induction

Question 1:

Q: Derive the expression for the induced EMF in a moving conductor using Faraday's law of induction.

Solution: Faraday's law of induction states that the induced EMF in a closed loop is equal to the rate of change of magnetic flux through the loop. For a moving conductor, the induced EMF is given by:

$$\text{EMF} = - \frac{d\Phi_B}{dt}$$

Where $\Phi_B$ is the magnetic flux through the conductor.

For a straight conductor of length $l$ moving with velocity $v$ perpendicular to a magnetic field $B$ , the area swept by the conductor in time $dt$ is $l \cdot v \cdot dt$ . Thus, the flux through this area is:

$$\Phi_B = B \cdot l \cdot v \cdot dt$$

Taking the derivative with respect to time:

$$\text{EMF} = - \frac{d\Phi_B}{dt} = - B \cdot l \cdot v$$

Thus, the induced EMF is $\text{EMF} = - B \cdot l \cdot v$ .

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Question 2:

Q: Explain Lenz's law and how it is consistent with the law of conservation of energy.

Solution: Lenz’s law states that the direction of the induced current (or EMF) is such that it opposes the change in magnetic flux that produced it. This law is a consequence of the principle of conservation of energy.

If the induced current did not oppose the change in flux, it would increase the flux even further, leading to a self-amplifying cycle. This would violate the conservation of energy, as it would create perpetual motion. Therefore, the induced current must oppose the flux change to ensure that energy is not created or destroyed in the process.

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Question 3:

Q: Derive the expression for the self-inductance of a solenoid.

Solution: The self-inductance $L$ of a solenoid is the ratio of the induced EMF to the rate of change of current through it.

For a solenoid with $N$ turns, length $l$ , and cross-sectional area $A$ , the magnetic flux $\Phi_B$ is given by:

$$\Phi_B = B \cdot A$$

Where $B = \mu_0 \frac{N}{l} I$ , $\mu_0$ is the permeability of free space, $N$ is the number of turns, and $I$ is the current.

Thus, the flux is:

$$\Phi_B = \mu_0 \frac{N}{l} I \cdot A$$

The induced EMF $\epsilon$ is:

$$\epsilon = -N \frac{d\Phi_B}{dt} = -N \frac{d}{dt} \left( \mu_0 \frac{N}{l} I \cdot A \right)$$

Simplifying:

$$\epsilon = - \mu_0 \frac{N^2 A}{l} \frac{dI}{dt}$$

The self-inductance $L$ is defined as $\epsilon = -L \frac{dI}{dt}$ , so:

$$L = \mu_0 \frac{N^2 A}{l}$$

This is the expression for the self-inductance of a solenoid.

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Question 4:

Q: A coil of $500$ turns and area $2 \, \text{cm}^2$ is placed in a magnetic field of $0.2 \, \text{T}$ . The magnetic field is increased to $0.4 \, \text{T}$ in $0.1 \, \text{s}$ . Calculate the induced EMF in the coil.

Solution: The induced EMF is given by Faraday's law:

$$\text{EMF} = - N \frac{d\Phi_B}{dt}$$

Where:

• $N = 500$ is the number of turns,
• $A = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2$ ,
• $B_1 = 0.2 \, \text{T}$ and $B_2 = 0.4 \, \text{T}$ are the initial and final magnetic fields,
• $\Delta t = 0.1 \, \text{s}$ .

The change in magnetic flux is:

$$\Delta \Phi_B = (B_2 - B_1) \cdot A = (0.4 - 0.2) \cdot 2 \times 10^{-4} = 0.2 \times 2 \times 10^{-4} = 4 \times 10^{-5} \, \text{Wb}$$

Now, using Faraday’s law:

$$\text{EMF} = - 500 \times \frac{4 \times 10^{-5}}{0.1} = - 500 \times 4 \times 10^{-4} = - 0.2 \, \text{V}$$

Thus, the induced EMF is $0.2 \, \text{V}$ .

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Question 5:

Q: Derive the expression for the energy stored in an inductor.

Solution: The energy $U$ stored in an inductor is given by the work done to establish the current in the inductor. The instantaneous power dissipated in the inductor is:

$$P = \text{EMF} \times I$$

From Lenz’s law, the induced EMF is $\text{EMF} = -L \frac{dI}{dt}$ . Thus, the instantaneous power is:

$$P = - L \frac{dI}{dt} \times I$$

The total work (or energy) stored in the inductor is the integral of power with respect to time:

$$U = \int_0^I P \, dt = \int_0^I -L I \, dI$$

Integrating:

$$U = \frac{1}{2} L I^2$$

Thus, the energy stored in the inductor is $U = \frac{1}{2} L I^2$ .

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Question 6:

Q: What is mutual inductance? Derive the expression for mutual inductance between two coils.

Solution: Mutual inductance $M$ is the property of two coils that measures how much the magnetic flux due to the current in one coil links with the second coil.

The mutual inductance between two coils is given by:

$$\text{EMF}_2 = - M \frac{dI_1}{dt}$$

Where:

• $M$ is the mutual inductance,
• $\frac{dI_1}{dt}$ is the rate of change of current in coil 1,
• $\text{EMF}_2$ is the induced EMF in coil 2.

The mutual inductance is defined as:

$$M = \frac{\Phi_{21}}{I_1}$$

Where $\Phi_{21}$ is the magnetic flux through coil 2 due to the current in coil 1.

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Question 7:

Q: A coil of resistance $10 \, \Omega$ is placed in a time-varying magnetic field. If the magnetic field changes from $0 \, \text{T}$ to $0.2 \, \text{T}$ in $0.05 \, \text{s}$ , and the area of the coil is $0.01 \, \text{m}^2$ , calculate the current induced in the coil.

Solution: The induced EMF is given by:

$$\text{EMF} = - \frac{d\Phi_B}{dt}$$

Where the change in magnetic flux is:

$$\Delta \Phi_B = (B_2 - B_1) \cdot A = (0.2 - 0) \cdot 0.01 = 0.002 \, \text{Wb}$$

The induced EMF is:

$$\text{EMF} = \frac{0.002}{0.05} = 0.04 \, \text{V}$$

Using Ohm’s law, the current $I$ is:

$$I = \frac{\text{EMF}}{R} = \frac{0.04}{10} = 0.004 \, \text{A}$$

Thus, the induced current is $0.004 \, \text{A}$ .

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Question 8:

Q: Discuss the phenomenon of eddy currents and derive the expression for the power loss due to eddy currents in a conductor.

Solution: Eddy currents are circulating currents induced in a conductor when there is a change in the magnetic field. These currents oppose the change in the magnetic flux and lead to energy dissipation in the form of heat.

The power loss due to eddy currents in a conductor is given by:

$$P = I_{\text{eddy}}^2 R$$

Where $I_{\text{eddy}}$ is the eddy current and $R$ is the resistance of the conductor. The induced current depends on the rate of change of magnetic flux and the resistivity of the material.

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Question 9:

Q: A solenoid has $1000$ turns, length $0.2 \, \text{m}$ , and cross-sectional area $1 \, \text{cm}^2$ . If the current in the solenoid changes from $2 \, \text{A}$ to $0 \, \text{A}$ in $0.1 \, \text{s}$ , calculate the induced EMF.

Solution: The self-inductance of the solenoid is given by:

$$L = \mu_0 \frac{N^2 A}{l}$$

Where:

• $N = 1000$ ,
• $A = 1 \times 10^{-4} \, \text{m}^2$ ,
• $l = 0.2 \, \text{m}$ ,
• $\mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A}$ .

So,

$$L = 4 \pi \times 10^{-7} \times \frac{1000^2 \times 1 \times 10^{-4}}{0.2} = 6.28 \, \text{H}$$

The induced EMF is:

$$\text{EMF} = -L \frac{dI}{dt}$$

Where:

$$\frac{dI}{dt} = \frac{0 - 2}{0.1} = -20 \, \text{A/s}$$

Thus,

$$\text{EMF} = -6.28 \times (-20) = 125.6 \, \text{V}$$

The induced EMF is $125.6 \, \text{V}$ .

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