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

\[\left( \frac{1}{10\pi} \right)\pi = \frac{1}{10}  s\]

\[e = \frac{d\phi}{dt} = \frac{2nBA}{dt}\]

\[     = \frac{2 \times 100 \times 4 \times {10}^{- 4} \times 25 \times {10}^{- 4}}{1/10}\]

\[     = 2 \times  {10}^{- 3}   V\]

(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 × `1/10`

= 5 × 10⁻⁵ C