A conducting loop of area 5.0 cm^{2} is placed in a magnetic field which varies sinusoidally with time as B = B_{0} sin ωt where B_{0} = 0.20 T and ω = 300 s^{−1}. The normal to the coil makes an angle of 60° with the field. Find (a) the maximum emf induced in the coil, (b) the emf induced at τ = (π/900)s and (c) the emf induced at t = (π/600) s.

#### Solution

Given:-

Area of the coil, A = 5 cm^{2} = 5 × 10^{−4} m^{2}

The magnetic field at time t is given by

B = B_{0} sin ωt = 0.2 sin (300t)

Angle of the normal of the coil with the magnetic field, θ = 60°

(a) The emf induced in the coil is given by

\[e = \frac{- d\theta}{dt} = \frac{d}{dt}(BA \cos \theta)\]

\[ = \frac{d}{dt}\left[ \left( B_0 \sin \omega t \right) \times 5 \times {10}^{- 4} \times 1/2 \right]\]

\[ = B_0 \times \frac{5}{2} \times {10}^{- 4} \frac{d}{dt}(\sin \omega t)\]

\[ = \frac{B_0 5}{2} {10}^{- 4} \omega\left( \cos \omega t \right)\]

\[ = \frac{0 . 2 \times 5}{2} \times 300 \times {10}^{- 4} \times \cos \omega t\]

\[ = 15 \times {10}^{- 3} cost \omega t\]

The induced emf becomes maximum when cos ωt becomes maximum, that is, 1.

Thus, the maximum value of the induced emf is given by

\[e_{max} = 15 \times {10}^{- 3} = 0 . 015 V\]

(b) The induced emf at t = \[\left( \frac{\pi}{900} \right) s\] is given by

e = 15 × 10^{−3} × cos ωt

= 15 × 10^{−3} × cos\[\left( 300\times\frac{\pi}{900} \right)\]

= 15 × 10^{−3} × `1/2`

\[= \frac{0 . 015}{2} = 0 . 0075 = 7 . 5 \times {10}^{- 3} V\]

(c) The induced emf at t = \[\frac{\pi}{600} s\] is given by

e = 15 × 10^{−3} × cos \[\left( 300 \times \frac{\pi}{600} \right)\]

= 15 × 10^{−3}^{ }× 0 = 0 V