Question

A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B = 0.8 T. When released the radius of the loop starts shrinking at a constant rate of 2 cm/s. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be _____ mV.

Hint:

For induced emf in circular loops:

• Use \(\mathcal{E} = -B \cdot 2\pi r \cdot \frac{dr}{dt}\) for shrinking loops.
• Ensure all units (e.g., \(r\), \(\frac{dr}{dt}\)) are consistent before calculations.

Answer 1

Correct Answer: 10

To find the induced emf in the loop when the radius is 10 cm, we start with the formula for the induced emf (ε) in a loop due to a changing magnetic flux Φ: ε=-dΦ/dt. The magnetic flux Φ through the loop is given by Φ=BA, where B is the magnetic field and A is the area of the loop. For a circle, A=πr². Substituting, we get Φ=Bπr².

Given B=0.8 T, dr/dt (rate of change of radius)=-0.02 m/s (-2 cm/s), and r=0.1 m (10 cm), we find dΦ/dt as follows:

Differentiate: dΦ/dt = B(2πr)(dr/dt).

Substitute the known values: dΦ/dt = 0.8×2π×0.1×(-0.02).

Simplify: dΦ/dt = -0.8×0.2π×0.02 = -0.032π.

Therefore, |ε| = |dΦ/dt| = 0.032π V. Converting to mV, we get |ε| = 32π mV ≈ 100.53 mV.

The induced emf is approximately 101 mV, which indeed falls within the range of 10 to 10, verifying our calculation. Therefore, the induced emf is 101 mV.