Q146 / 180 · COMEDK 2024 (Evening Shift) · Physics
A conducting circular loop is placed in a uniform magnetic field $\mathrm{B}=0.125 \mathrm{~T}$ with its plane perpendicular to the loop. If the radius of the loop is made to shrink at a constant rate of $2 \mathrm{~mm} \mathrm{~s}^{-1}$, then the induced emf when the radius is $4 \mathrm{~cm}$ is
- A$\dfrac{2}{3}\mu V$
- Correct answerB$20 \pi \mu V$
- C$0.52 \pi \mu V$
- D$\dfrac{3 \pi}{2} \mu V$
Solution
The magnetic flux $\Phi$ through the circular loop is given by $\Phi = B A = B (\pi r^2)$, where $B$ is the magnetic field and $r$ is the radius of the loop.
The induced electromotive force (emf) $\varepsilon$ is given by Faraday's law of induction: $\varepsilon = -\dfrac{d\Phi}{dt}$.
Since $B$ is constant, $\varepsilon = -\dfrac{d}{dt}(B \pi r^2) = -B \pi (2r \dfrac{dr}{dt}) = -2 \pi B r \dfrac{dr}{dt}$.
Given $B = 0.125 \text{ T}$, $r = 4 \text{ cm} = 0.04 \text{ m}$, and the rate of shrinking $\dfrac{dr}{dt} = -2 \text{ mm s}^{-1} = -2 \times 10^{-3} \text{ m s}^{-1}$.
Substituting these values into the expression for magnitude of induced emf:
$|\varepsilon| = 2 \pi (0.125) (0.04) (2 \times 10^{-3})$
$|\varepsilon| = 2 \pi (\dfrac{1}{8}) (4 \times 10^{-2}) (2 \times 10^{-3})$
$|\varepsilon| = 2 \pi (\dfrac{1}{8}) (8 \times 10^{-5}) = 2 \pi \times 10^{-5} \text{ V}$.
Converting to microvolts ($\mu V$):
$|\varepsilon| = 2 \pi \times 10^{-5} \times 10^{6} \mu V = 20 \pi \mu V$.
Answer: $20 \pi \mu V$
The induced electromotive force (emf) $\varepsilon$ is given by Faraday's law of induction: $\varepsilon = -\dfrac{d\Phi}{dt}$.
Since $B$ is constant, $\varepsilon = -\dfrac{d}{dt}(B \pi r^2) = -B \pi (2r \dfrac{dr}{dt}) = -2 \pi B r \dfrac{dr}{dt}$.
Given $B = 0.125 \text{ T}$, $r = 4 \text{ cm} = 0.04 \text{ m}$, and the rate of shrinking $\dfrac{dr}{dt} = -2 \text{ mm s}^{-1} = -2 \times 10^{-3} \text{ m s}^{-1}$.
Substituting these values into the expression for magnitude of induced emf:
$|\varepsilon| = 2 \pi (0.125) (0.04) (2 \times 10^{-3})$
$|\varepsilon| = 2 \pi (\dfrac{1}{8}) (4 \times 10^{-2}) (2 \times 10^{-3})$
$|\varepsilon| = 2 \pi (\dfrac{1}{8}) (8 \times 10^{-5}) = 2 \pi \times 10^{-5} \text{ V}$.
Converting to microvolts ($\mu V$):
$|\varepsilon| = 2 \pi \times 10^{-5} \times 10^{6} \mu V = 20 \pi \mu V$.
Answer: $20 \pi \mu V$
Hence, the correct answer is option: 'B'.