The homogeneous magnetic field is perpendicular to the plane of the drawing and changes in time according to the law $B=kt$. The resistance per unit length of the conductor in r. Correctly match the modulus of current in the wire mentioned in column I with the values given in column II.

	Column-I		Column-II
a)	AB	p)	$\frac{3}{11}\frac{ka}{r}$
b)	BC	q)	$\frac{1}{22}\frac{ka}{r}$
c)	EF	r)	$\frac{7}{22}\frac{ka}{r}$
d)	CD	s)	$\frac{13}{22}\frac{ka}{r}$

$\epsilon =\frac{-d\phi }{dt}$

The emf induced in the loop ABCD is ka². emf induced in the loop BCEF is $\frac{{\mathrm{ka}}^2}{2}$.

Let i₁ be the current in ABCD and it will be in anti-clock wise direction, and i₂ be the current in BCEF in anti-clock wise direction.

in ABCD ${\mathrm{ka}}^2-3{\mathrm{i}}_1\mathrm{ar}-({\mathrm{i}}_1-{\mathrm{i}}_2)\mathrm{ar}=0.....\left(1\right)$

in BCEF $\frac{{\mathrm{ka}}^2}{2}-2{\mathrm{i}}_2\mathrm{ra}-({\mathrm{i}}_1-{\mathrm{i}}_2)\mathrm{ra}=0....\left(2\right)$

on solving (1) and (2), we get 

${\mathrm{i}}_1=\frac{7\mathrm{ka}}{22\mathrm{r}}, {\mathrm{i}}_2=\frac{3\mathrm{ka}}{11\mathrm{r}}$.

So the current through BC is $\frac{\mathrm{ka}}{22\mathrm{r}}$