Two concentric coplanar circular loops of radii $r_{1}$ and $r_{2}$ carry currents of respectively $i_{1}$ and $i_{2}$ in opposite directions (one clockwise and the other anticlockwise.) The magnetic induction at the centre of the loops is half that due to $i_{1}$ alone at the centre. If $r_{2} = 2r_{1}$. The value of $i_{2}/i_{1}$ is

Question

Dr. Abubacker · Accepted Answer

(d) Magnetic field at centre due to smaller loop
$B_{1} = \frac{\mu_{0}}{4\pi}.\frac{2\pi i_{1}}{r_{1}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Due to Bigger loop $B_{2} = \frac{\mu_{0}}{4\pi}.\frac{2\pi i_{2}}{r_{2}}$. So net magnetic field at centre
$B = B_{1} - B_{2} = \frac{\mu_{0}}{4\pi} 	imes 2\pi\left( \frac{i_{1}}{r_{1}} - \frac{i_{2}}{r_{2}} ight)$
According to question $B = \frac{1}{2} 	imes B_{1}$
$\Rightarrow \frac{\mu_{0}}{4\pi}.2\pi\left( \frac{i_{1}}{r_{1}} - \frac{i_{2}}{r_{2}} ight) = \frac{1}{2} 	imes \frac{\mu_{0}}{4\pi}.\frac{2\pi i_{1}}{r_{1}}$
$\frac{i_{1}}{r_{1}} - \frac{i_{2}}{r_{2}} = \frac{i_{1}}{2r_{1}} \Rightarrow \frac{i_{1}}{2r_{1}} = \frac{i_{2}}{r_{2}} \Rightarrow \frac{i_{1}}{i_{2}} = 1\ \ \ \lbrack r_{2} = 2r_{1}brack$

Explanation

Reviewed by Dr. Abubacker

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