A conducting loop of radius $10/[\pi]^{1/2}$ cm is placed perpendicular to a uniform magnetic field $0.5$ T. The magnetic field is decreased to zero in $0.5$ s at a steady rate. The induced emf in the circular loop at $0.25$ s is

Given radius of the loop, $r = \dfrac{10}{\sqrt{\pi}}\text{ cm} = \dfrac{0.1}{\sqrt{\pi}}\text{ m}$

Area of the loop, $A = \pi r^2 = \pi \left(\dfrac{0.1}{\sqrt{\pi}}\right)^2 = 0.01\text{ m}^2$

Initial magnetic field, $B_i = 0.5\text{ T}$

Final magnetic field, $B_f = 0\text{ T}$

Time interval, $\Delta t = 0.5\text{ s}$

Since the magnetic field decreases at a steady rate, the rate of change of magnetic field is constant.

$\dfrac{dB}{dt} = \dfrac{B_f - B_i}{\Delta t} = \dfrac{0 - 0.5}{0.5} = -1\text{ T/s}$

According to Faraday's law of induction, the induced emf is given by:

$e = -\dfrac{d\Phi}{dt} = -A \dfrac{dB}{dt}$

$e = -0.01 \times (-1) = 0.01\text{ V} = 10\text{ mV}$

Since the rate of change of magnetic field is constant, the induced emf is constant at all times during the interval. Thus, the induced emf at $t = 0.25\text{ s}$ is $10\text{ mV}$.

Answer: $10\text{ mV}$