Q128 / 200 · NEET 2010 (Screening) · Physics

A conducting circular loop is placed in a uniform magnetic field, $B=0.025 \mathrm{~T}$ with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of $1 \mathrm{mms}^{-1}$. The induced emf when the radius is $2 \mathrm{~cm}$, is

  • A
    $2 \pi \mu \mathrm{V}$
  • Correct answer
    B
    $\pi \mu \mathrm{V}$
  • C
    $\frac{\pi}{2} \mu \mathrm{V}$
  • D
    $2 \mu \mathrm{V}$

Solution

Given: The magnetic field in the region is, \(\mathrm{B}=0.025 \mathrm{~T}\). Radius of the loop, \(\mathrm{r}=2 \mathrm{~cm}=2 \times 10^{-2}\) Constant rate at which radius of the loop shrimks, \(\frac{\mathrm{dr}}{\mathrm{dt}}=1 \times 10^{-3} \mathrm{~ms}^{-1}\) To find: The induced emf in the loop when radius is 2 cm. The magnetic flux linked with the loop is given by: \(\begin{aligned} & \phi=\mathrm{BA} \cos \theta \\ & =\mathrm{B}\left(\pi \mathrm{r}^2\right) \cos \mathrm{O}^{\circ} \\ & \Rightarrow \mathrm{B} \pi \mathrm{r}^2 \end{aligned}\) The magnitude of induced emf is given as: \(\begin{aligned} & |\epsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}} \\ & =\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B} \pi \mathrm{r}^2\right) \\ & =\mathrm{B} \pi 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \end{aligned}\) Substitute the values. \(\begin{aligned} & |\epsilon|=0.025 \times \pi \times 2 \times 2 \times 10^{-2} \times 1 \times 10^{-3} \\ & =\pi \times 10^{-6} \mathrm{~V} \\ & \Rightarrow \pi \mu \mathrm{V} \end{aligned}\)

Hence, the correct answer is option: 'B'.

Relevant Questions