Question

A conducting loop of area 5.0 cm² is placed in a magnetic field which varies sinusoidally with time as B = B₀ sin ωt where B₀ = 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.

Answer 1

Given:
Area of the coil, A = 5 cm² = 5 × 10⁻⁴ m²
The magnetic field at time t is given by
B = B₀ 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
$$\begin{gathered} e=\frac{-d\theta }{dt}=\frac{d}{dt}(BA\cos \theta ) \\ =\frac{d}{dt}\left[\left(B_0\sin \mathrm{\omega }t\right)\times 5\times {10}^{-4}\times 1/2\right] \\ =B_0\times \frac{5}{2}\times {10}^{-4}\frac{d}{dt}(\sin \mathrm{\omega }t) \\ =\frac{B_{0}5}{2}{10}^{-4}\omega \left(\cos \mathrm{\omega }t\right) \\ =\frac{0.2\times 5}{2}\times 300\times {10}^{-4}\times \cos \mathrm{\omega }t \\ =15\times {10}^{-3}\mathrm{cost}\mathrm{\omega }t \end{gathered}$$

The induced emf becomes maximum when cos ωt becomes maximum, that is, 1.
Thus, the maximum value of the induced emf is given by
$e_{max}=15\times {10}^{-3}=0.015\mathrm{V}$
(b) The induced emf at t = $\left(\frac{\mathrm{\pi }}{900}\right)\mathrm{s}$ is given by
e = 15 × 10⁻³ × cos ωt
= 15 × 10⁻³ × cos $\left(300\times \frac{\mathrm{\pi }}{900}\right)$
= 15 × 10⁻³ × $\frac{1}{2}$
$=\frac{0.015}{2}=0.0075=7.5\times {10}^{-3}\mathrm{V}$
(c) The induced emf at t = $\frac{\mathrm{\pi }}{600}\mathrm{s}$ is given by
e = 15 × 10⁻³ × cos $\left(300\times \frac{\mathrm{\pi }}{600}\right)$
= 15 × 10⁻³ × 0 = 0 V