Question

A magnetic field of $2 \times 10^{-2}\,\text{T}$ acts at right angles to a coil of area $100\,\text{cm}^2$ with 50 turns. The average e.m.f. induced in the coil is $0.1\,\text{V}$ when it is removed from the field in time $t$. The value of $t$ is

• A. $2 \times 10^{-3}$
• B. 0.5 s
• C. 0.1 s
• D. 1 s

Hint:

Physics Tip : Induced emf only exists while the magnetic flux is changing. Faster removal results in a higher induced voltage.

Answer 1

The Correct Option is C

Concept: Physics (Electromagnetic Induction) - Faraday's Law of Induction.

Step 1: Identify the initial and final magnetic flux. Initial flux $\phi_1 = NBA$ (since it acts at right angles). Final flux $\phi_2 = 0$ (once removed from the field).
• $N = 50$ turns
• $B = 2 \times 10^{-2}$ T
• $A = 100\text{ cm}^2 = 10^{-2}\text{ m}^2$

Step 2: Apply Faraday's Law. The induced emf $e$ is given by the rate of change of flux1093]: $$e = -\frac{\Delta\phi}{t} = \frac{\phi_1 - \phi_2}{t} = \frac{NBA}{t}$$

Step 3: Substitute and solve for $t$. $$0.1 = \frac{50 \times (2 \times 10^{-2}) \times 10^{-2}}{t} \text{ }$$ $$0.1 = \frac{10^{-2}}{t} \implies t = \frac{0.01}{0.1} = 0.1 \text{ s}$$ $$ \therefore \text{The value of 't' is 0.1 s.} \text{ } $$