Question

The inner loop has an area of $5\times {10}^{-4}{\text{m}}^2$ and a resistance of $2\Omega$ (see figure). The larger circular loop is fixed and has a radius of $0.1m$. Both the loops are concentric and coplanar. The smaller loop is rotated with an angular velocity of $\omega rad{s}^{-1}$ about its diameter. The magnetic flux linked with the smaller loop is

• A. $2\pi \times {10}^{-6}Wb$
• B. $\pi \times {10}^{-9}Wb$
• C. $\pi \times {10}^{-9}\cos \omega tWb$
• D. Zero

Answer 1

The correct option is C $\pi \times {10}^{-9}\cos \omega tWb$

As flux, $\varphi =B.A=\mid B_l\mid \mid A_s\mid cos\theta$
where $A_s=$ area of smaller loop
$B_l=$ Magnetic field due to larger loop
Since angular velocity, $\omega =\frac{\theta }{t}$
$\theta =\omega t$
Therefore flux, $\begin{array}{}\varphi =\mid B_l\mid \mid A_s\mid cos\omega t& \text{(1)}\end{array}$
Magnetic field at centre of current carrying loop, $B=\frac{{\mu }_{o}I}{2R}$
where $R$ is the radius of the loop

Substituting the value in $\left(1\right)$,
$\varphi =\frac{{\mu }_{o}I}{2R}\mid A_s\mid cos\omega t$
$\varphi =\frac{4\pi \times {10}^{-7}\times 1}{2\times 0.1}\times (5\times {10}^{-4})cos\omega t$
$\Rightarrow \varphi =\pi \times {10}^{-9}cos\omega tWb$