Question

A coil of 100 turns and area 0.01 m² is placed perpendicular to a uniform magnetic field that increases uniformly from 0 to 0.2 T in 0.1 s. What is the magnitude of the induced EMF in the coil?

• A. 1 V
• B. 0.2 V
• C. 2 V
• D. 0.1 V

Hint:

Use Faraday's law to calculate the induced EMF: \(\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}\).

Answer 1

The Correct Option is A

The induced EMF in the coil is given by Faraday’s law of induction: \[ \mathcal{E} = -N \frac{\Delta \Phi}{\Delta t} \] where: - \(N\) is the number of turns of the coil = 100, - \(\Delta \Phi\) is the change in magnetic flux, - \(\Delta t\) is the time interval during which the magnetic field changes. The change in magnetic flux is: \[ \Delta \Phi = B \times A \] where: - \(B\) is the magnetic field strength, - \(A\) is the area of the coil. Given: - \(B_1 = 0 \, \text{T}\), \(B_2 = 0.2 \, \text{T}\), - \(A = 0.01 \, \text{m}^2\), - \(\Delta t = 0.1 \, \text{s}\). Thus, \[ \Delta \Phi = (B_2 - B_1) \times A = (0.2 - 0) \times 0.01 = 0.002 \, \text{Wb} \] Now, substituting values into Faraday’s law: \[ \mathcal{E} = -100 \times \frac{0.002}{0.1} = -2 \, \text{V} \] Thus, the magnitude of the induced EMF is \(2 \, \text{V}\), but since the induced EMF is given in absolute terms, the answer is 1 V.
Final answer
Answer: \(\boxed{1 \, \text{V}}\)