Q92 / 100 · BITSAT 2023 (Memory Based Paper 1) · Physics
A conducting loop of radius $\frac{10}{\sqrt{\pi}} \mathrm{cm}$ is placed perpendicular to a uniform magnetic field of $0.5 T$. 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:
- A$\mathrm{emf}=1 \mathrm{mV}$
- Correct answerB$\mathrm{emf}=10 \mathrm{mV}$
- C$\mathrm{emf}=100 \mathrm{mV}$
- D$\mathrm{emf}=5 \mathrm{mV}$
Solution
As $\left.\varepsilon\right|_{t=0.5 \mathrm{sec}}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}$
$=-\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}} \quad\left[\because \theta=0^{\circ} \Rightarrow \cos \theta=1\right]$
$=-\pi \times\left(\frac{10}{\sqrt{\pi}}\right)^2 \times 10^{-4} \times \frac{0-0.5}{0.5}=10^{-2} \mathrm{~V}=10 \mathrm{mV}$
As $\frac{d B}{d t}=$ constant $\Rightarrow$ Induced emf will not change with time. So, $\left.\mathrm{e}\right|_{0,5 \mathrm{sec}}=\left.\mathrm{e}\right|_{0.25 \mathrm{sec}}=10 \mathrm{mV}$
$=-\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}} \quad\left[\because \theta=0^{\circ} \Rightarrow \cos \theta=1\right]$
$=-\pi \times\left(\frac{10}{\sqrt{\pi}}\right)^2 \times 10^{-4} \times \frac{0-0.5}{0.5}=10^{-2} \mathrm{~V}=10 \mathrm{mV}$
As $\frac{d B}{d t}=$ constant $\Rightarrow$ Induced emf will not change with time. So, $\left.\mathrm{e}\right|_{0,5 \mathrm{sec}}=\left.\mathrm{e}\right|_{0.25 \mathrm{sec}}=10 \mathrm{mV}$
Hence, the correct answer is option: 'B'.