Practicing Success

Target Exam



-- Physics - Section A


Electromagnetic induction and alternating currents


A wire ring of radius R is fixed in a horizontal plane. The wire of the ring has a resistance of $λΩm^{–1}$. There is a uniform vertical magnetic field B in entire space. A perfectly conducting rod ($l$) is kept along the diameter of the ring. The rod is made to move with a constant acceleration a in a direction perpendicular to its own length. What is the current through the rod at the instant it has travelled through a distance $x=\frac{R}{2}$ ?


$\frac{16B}{4\pi }\frac{\sqrt{3aR}}{λ}$

$\frac{18B}{5\pi }\frac{\sqrt{3aR}}{λ}$

$\frac{18B}{5\pi }\frac{\sqrt{2aR}}{λ}$

$\frac{16B}{4\pi }\frac{\sqrt{2aR}}{λ}$

Correct Answer:

$\frac{18B}{5\pi }\frac{\sqrt{3aR}}{λ}$


Speed of the rod when it has travelled a distance $x=\frac{R}{2}$ is

$v= \sqrt{2ax}= \sqrt{aR}$

Length of rod inside the ring $L= AC= 2\sqrt{R^2-\frac{R^2}{4}}= \sqrt{3}R$

 Emf induced in the rod at this instant is

$E= BvL= B. \sqrt{aR} \sqrt{3}. R= \sqrt{3a} B. R^{3/2}$

$sin \theta = \frac{x}{R}=\frac{1}{2}⇒ \theta = 30°$

∴ Length of arc $ABC= \frac{2\pi R}{6}=\frac{\pi R}{3}$

Length of arc $ADC= \frac{5}{6}. 2\pi R = \frac{5\pi R}{3}$

Resistance of arc ABC; $r_1=\frac{\pi R}{3}λ$

Resistance of arc ADC; $r_2= \frac{5\pi R}{2}λ$

Equivalent resistance $= r= \frac{r_1r_2}{r_1+r_2}= \frac{5\pi }{18} λR$

∴ Current through the rod $I=\frac{E}{r}=\frac{18B}{5\pi }\frac{\sqrt{3aR}}{λ}$