Question

Figure shows a uniform circular loop of radius $a$ having resistance per unit length $\rho$ placed in a unifrom magnetic field $B$ perpendicular to plane of figure. A uniform rod of length $2a$ & resistance $R$ moves with a velocity $v$ as shown.
Find the current in the rod when it has moved a distance $\frac{a}{2}$ from the centre of circular loop.
$\text{Given B}=\sqrt{3},\text{a}=2,\text{v}=3,\rho =\frac{9}{4\pi },\text{R}=\frac{2}{\sqrt{3}}$ all in SI units.

Answer 1

Using Pythogaras theorem, we can calculate the length rod on the circular loop.
The half of the length of the rod on the loop is given by,
$x^2=a^2-(\frac{a}{2}{)}^2\Rightarrow x=\frac{\sqrt{3}a}{2}$
The total length of the rod on the circular loop is $2x=\sqrt{3}a$

The motional emf of the rod is given by $ϵ=Blv=B\times \sqrt{3}a\times v$
Substituting the values we get, $ϵ=18V$

Let $\theta$ be the angle subtented by one end of the rod and the center of the rod at the center of the circular loop.
$cos\theta =\frac{1}{2}$
$\theta ={60}^{\circ }$
The total angle made by the rod at the center of the circular loop is ${120}^{\circ }$
$\Rightarrow$ one third of the circular loop is to the right of the rod and two thirds is on the left of the rod.
$\Rightarrow$ their resistances are also in the ratio 1:2

The circumference of the circular loop is $2\pi a=4\pi$
Given resistance per unit length of the loop $\frac{9}{4\pi }$
$\therefore$ the resistance of the whole loop is $\frac{9}{4\pi }\times 4\pi =9\Omega$

Now dividing this resistance in the ratio 1:2 (for the right and the left parts of the loop w.r.t the rod) we have $3\Omega$ for the right part and $6\Omega$ for the left part.

And the resistance of the length of the rod that is on the circular loop is $1\Omega$

Now the whole system can be written as shown in the figure.


$6\Omega$ is parallel to $3\Omega$ and the this combination is parallel to $1\Omega$.

Their effective resistance is $3\Omega$.
The current through the rod is $i=\frac{18}{3}=6A$