A circular coil of radius 10 cm is placed in a uniform magnetic field of 3.0 × 10 –5 T with its plane perpendicular to the field initially. It is rotated at constant angular speed about an axis along the diameter of coil and perpendicular to magnetic field so that it undergoes half of rotation in 0.2 s. The maximum value of EMF induced (in μV) in the coil will be close to the integer _______.

Solution

Physics sir · 4 steps

Half a turn is π radians, so get ω from that first; the peak EMF is then just BAω.

1. Flux as it turns. With the coil's normal at angle ω t to the field, the flux is Φ = BA cos ω t. Then e = -dΦ/dt = BAω sin ω t, so the largest EMF is BAω.

   eₘₐₓ = BAω

2. Find omega and area. Half a rotation is π rad in 0.2 s, so ω = π/0.2 = 5π rad s⁻¹. The area is A = π r² = π × (0.1)² = 0.01π m².
3. Put in numbers. eₘₐₓ = 3.0 × 10⁻⁵ × 0.01π × 5π = 1.5 × 10⁻⁶ × π² ≈ 1.48 × 10⁻⁵ V.
4. So. eₘₐₓ ≈ 14.8 μV, which is closest to 15.

The common slip: 30. Treating 0.2 s as the time for a full rotation gives ω = 10π rad s⁻¹ and about 30, not 15.

The idea to keep. For a coil spinning in a field, the peak EMF is NBAω. Turn the given time into ω carefully: half a turn is π radians, a full turn is 2π.

Answer from the NTA final answer key. Solution by PrepWiser.