First slide

Faraday's law of Electromagnetic induction

Question

A coil of copper having 1O0O turns is placed in a magnetic field (B = 4x 10^-3 T) perpendicular to its axis. The cross-sectional area of the coil is 0.05 m². If it turns through 180^o in 0.01 sec, then the e.m.f. induced in the coil is

Easy

A

0.4 V
B

0.2 V
C

40 V
D

4 V

Solution

The magnetic flux $ϕ_{1}$ passing through a coil when it is placed perpendicular to magnetic field B is
$Φ = nBA$
When the coil is rotated through 90^o
$\begin{array}{l} Φ_{2} = - nBA \\ ∴ ΔΦ = Φ_{2} - Φ_{1} \\ = - nBA - nBA = - 2 nBA \\ Now e = - \frac{ΔΦ}{Δt} = \frac{+ 2 nBA}{Δt} \\ = \frac{2 \times 1000 \times (4 \times 10^{- 3}) \times 0 \cdot 05}{0 \cdot 01} = 40 V \end{array}$

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