Question

Two concentric, co-planar, circular loops have diameters $10\text{cm}$ and $1\text{m}$ and resistance per unit length of the wire, ${10}^{-4}\Omega /\text{m}$. A time-dependent voltage $V=(4+2.5t)\text{volts}$ is applied to the larger ring as shown. The current induced in the smaller loop is approximately,

• A. $1\text{A}$
• B. $2\text{A}$
• C. $1.25\text{A}$
• D. $2.5\text{A}$

Answer 1

The correct option is C $1.25\text{A}$

Given:
$r_1=1\text{m},r_2={10}^{-1}\text{m}$

$\therefore$ Resistance of outer loop,

$R_1=2\pi r_1\times {10}^{-4}=2\pi \times {10}^{-4}\Omega$

Resistance of inner loop,

$R_2=2\pi r_2\times {10}^{-4}=0.2\pi \times {10}^{-4}\Omega$

$\therefore$ current in the outer loop:

$I=\frac{V}{R_1}=\frac{(4+2.5t)}{2\pi \times {10}^{-4}}$

Magnetic field at the common center of the loops is,
$B=\frac{{\mu }_{0}I}{2r_1}$

$B=\frac{4\pi \times {10}^{-7}}{2\times 1}\times \frac{(4+2.5t)}{2\pi \times {10}^{-4}}$

$\Rightarrow B=2\times {10}^{-3}(2+1.25t)\text{T}$

Since, $r_2<<r_1$, the magnetic field throughout the inner loop can be assumed to be constant in magnitude, and its direction is uniform throughout.ie normally inwards.

Hence, flux linked with inner loop will be,

$\varphi =BA=2\times {10}^{-3}(2+1.25t)\times \pi (0.1{)}^2$

$\varphi =2\pi \times {10}^{-5}(2+1.25t)\text{Wb}$

$\therefore$ emf induced in smaller loop,

$e=\left|\frac{d\varphi }{dt}\right|=2\pi \times {10}^{-5}\frac{d(2+1.25t)}{dt}$

$\Rightarrow e=2.5\pi \times {10}^{-5}\text{V}$

Hence, the current in the inner loop is,

$i=\frac{|e|}{R_2}=\frac{2.5\pi \times {10}^{-5}}{0.2\pi \times {10}^{-4}}=1.25\text{A}$

Hence, option $\left(C\right)$ is the correct answer.