Question

A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular to the magnetic field. Somehow, the radus of the loop starts expanding at a constant rate of 1 mm/s. The magnitude of induced emf in the loop at an instant when the radius of the loop is 2 cm will be ___μV

Answer 1

Correct Answer: 50

Solution:
Given:

$B = 0.4$ T

$\frac{dr}{dt} = 1 \text{ mm/s} = 1 \times 10^{-3} \text{ m/s}$

$r = 2 \text{ cm} = 2 \times 10^{-2} \text{ m}$

$\theta = 0^\circ$ (plane perpendicular to the field)

The area of the loop is $A = \pi r^2$.
The magnetic flux is $\Phi = B \pi r^2 \cos 0^\circ = B \pi r^2$. Induced emf: $$ \varepsilon = -\frac{d(B \pi r^2)}{dt} = -B \pi \frac{d(r^2)}{dt} = -B \pi (2r \frac{dr}{dt}) = -2 \pi B r \frac{dr}{dt} $$ Substituting the given values: $$ \varepsilon = -2 \pi (0.4) (2 \times 10^{-2}) (1 \times 10^{-3}) = -1.6 \pi \times 10^{-5} \text{ V} $$ $$ \varepsilon = -16 \pi \times 10^{-6} \text{ V} = -16 \pi \mu \text{V} $$ Magnitude of the induced emf: $$ |\varepsilon| = 16 \pi \mu \text{V} $$ $$ |\varepsilon| = 16 \times 3.14159 \mu \text{V} \approx 50.265 \mu \text{V} $$ Therefore, the magnitude of the induced emf in the loop is approximately 50.265 $\mu V$.

 Answer: 50.265