Question

Find the induced emf in the loop when it's radius is 2 cm in a conducting circular loop with its plane perpendicular to the direction of magnetic field is placed in a uniform magnetic field, B = 0.025 T. The radius of the loop is made to shrink at a constant rate of 1 mm s^-1. .

• A. $2 \, \pi μ \text{V}$
• B. $\pi \mu \text{V}$
• C. 1 $\mu \text{V}$
• D. $2 \, μ \text{V}$

Answer 1

Answer: (B)

Here,
Magnetic field, B = 0.025 T
Radius of the loop, $\text{r} = \text{2 cm} = 2 \times 1 0^{- 2} m$
Constant rate at which radius of the loop shrinks,
$\frac{\text{dr}}{\text{dt}} = 1 \times 1 0^{- 3} \text{m s}^{- 1}$
Magnetic flux linked with the loop is
$\phi = \text{BA} \text{cos} \theta = \text{B} \left(\pi \left(\text{r}\right)^{2}\right) \text{cos} 0^{^\circ } = \text{B} \pi \left(\text{r}\right)^{2}$
From Faraday's law, the magnitude of the induced emf is
$\left|\epsilon \right| = \frac{\text{d} \phi}{\text{dt}} = \frac{\text{d}}{\text{dt}} \left(\text{B} \pi \left(\text{r}\right)^{2}\right) = \text{B} \pi 2 \text{r} \frac{\text{dr}}{\text{dt}}$
$\Rightarrow \left|\right. \epsilon \left|\right.$ $= \text{0.025} \times \pi \times 2 \times 2 \times 1 0^{- 2} \times 1 \times 1 0^{- 3}$
$\Rightarrow \left|\right. \epsilon \left|\right.$ $= \pi \times 1 0^{- 6} \text{ V} = \pi \, μ \text{V}$