Question

A closed coil having 100 turns is rotated in a uniform magnetic field B = 4.0 × 10⁻⁴ T about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm² and its resistance is 4.0 Ω. Find (a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field, (b) the average emf in a full turn and (c) the net charge displaced in part (a).

Answer 1

Given:
Number of turns in the coil, n = 100 turns
Magnetic field, B = 4 × 10⁻⁴
Area of the loop, A = 25 cm² = 25 × 10⁻⁴ m²

(a) When the coil is perpendicular to the field:
ϕ₁ = nBA
When the coil goes through the half turn:
ϕ₂ = nBA cos 180° = −nBA
∴ Δϕ = 2nBA
When the coil undergoes 300 revolutions in 1 minute, the angle swept by the coil is
300 × 2π rad/min = 10π rad/s
10π rad is swept in 1 s.
π rad is swept in $\left(\frac{1}{10\mathrm{\pi }}\right)\mathrm{\pi }=\frac{1}{10}\mathrm{s}$
$$\begin{gathered} e=\frac{d\mathrm{\varphi }}{dt}=\frac{2nBA}{dt} \\ =\frac{2\times 100\times 4\times {10}^{-4}\times 25\times {10}^{-4}}{1/10} \\ =2\times {10}^{-3}\mathrm{V} \end{gathered}$$

(b) ϕ₁ = nBA, ϕ₂ = nBA (θ = 360°)
Δϕ = 0, thus emf induced will be zero.

(c) The current flowing in the coil is given by
$i=\frac{e}{R}=\frac{2\times {10}^{-3}}{4}=\frac{1}{2}\times {10}^{-3}$
= 0.5 × 10⁻³ = 5 × 10⁻⁴ A
Hence, the net charge is given by
Q = idt = 5 × 10⁻⁴ × $\frac{1}{10}$
= 5 × 10⁻⁵ C