A conducting circular loop is placed in a uniform magnetic field, $B=0.025 \mathrm{~T}$ with its plane…

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
  1. $2 \pi \mu \mathrm{V}$
  2. $\pi \mu \mathrm{V}$
  3. $\frac{\pi}{2} \mu \mathrm{V}$
  4. $2 \mu \mathrm{V}$

Solution

Given:
Magnetic field, $B = 0.025 \, \text{T}$
Rate of shrinking of radius, $\frac{dr}{dt} = -1 \, \text{mm/s} = -1 \times 10^{-3} \, \text{m/s}$ (negative sign indicates shrinking)
Radius of the loop, $r = 2 \, \text{cm} = 0.02 \, \text{m}$
The magnetic flux $\Phi$ through the loop is given by:
$\Phi = B \cdot A$
Since the plane of the loop is perpendicular to the magnetic field, the angle between the magnetic field and the area vector is $0^\circ$, so $\cos 0^\circ = 1$.
For a circular loop, the area $A = \pi r^2$.
Therefore, $\Phi = B \pi r^2$
According to Faraday's Law of Electromagnetic Induction, the induced emf $\varepsilon$ is given by:
$\varepsilon = -\frac{d\Phi}{dt}$
$\varepsilon = -\frac{d}{dt}(B \pi r^2)$
Since $B$ and $\pi$ are constants, we can write:
$\varepsilon = -B \pi \frac{d}{dt}(r^2)$
Using the chain rule, $\frac{d}{dt}(r^2) = 2r \frac{dr}{dt}$
So, $\varepsilon = -B \pi (2r \frac{dr}{dt})$
$\varepsilon = -2 \pi B r \frac{dr}{dt}$
Now, substitute the given values:
$\varepsilon = -2 \pi (0.025 \, \text{T}) (0.02 \, \text{m}) (-1 \times 10^{-3} \, \text{m/s})$
$\varepsilon = 2 \pi \times 0.025 \times 0.02 \times 1 \times 10^{-3}$
$\varepsilon = 2 \pi \times 0.0005 \times 10^{-3}$
$\varepsilon = \pi \times 0.001 \times 10^{-3}$
$\varepsilon = \pi \times 10^{-6} \, \text{V}$
$\varepsilon = \pi \, \mu \text{V}$
The induced emf when the radius is $2 \, \text{cm}$ is $\pi \, \mu \text{V}$.

Asked in: NEET 2010 (Screening)

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