Question

Figure shows an equilateral triangular frame of side L. Frame starts entering into the region of uniform magnetic field B with constant velocity v. If i is instantaneous current through the frame, then the correct graph is. Take anticlockwise current direction to be positive.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B


$|i|=\frac{1}{R}\frac{d\varphi }{dt}$
$\varphi =BA\Rightarrow \frac{d\varphi }{dt}=B\frac{dA}{dt}$
$A=\frac{1}{2}b\times h=h^2\tan \theta$
$\frac{dA}{dt}=2h\tan \theta \frac{dh}{dt}$
Given that, $\frac{dh}{dt}$ is constant, since the traingle is moving at constant speed.
$\therefore \frac{dA}{dt}\propto h$
$\therefore i\propto \frac{d\varphi }{dt}\propto \frac{dA}{dt}\propto h$
So when the triangle is entering, current $i\propto h$, where h is the height of the triangle inside the field. Similarly, when the triangle is leaving, current $i\propto h$, where h is the height of the triangle outside of the field.