Question

A copper rod of length $l$ is rotated about the end perpendicular to the uniform magnetic field $B$ with constant angular velocity $\omega$. The induced e.m.f. between the two ends is

• A. $2 B \omega l^2$
• B. $B \omega l^2$
• C. $\frac{1}{2} B \omega l^2$
• D. $\frac{1}{4} B \omega l^2$

Answer 1

Answer: (C)

$\frac{1}{2} B \omega l^2$

Answer 2

To calculate the em f we can imagine a closed loop by connecting the centre with any point on the circumference, say $X$ with a resistor

Potential difference across the resistor is then equal to the induced emf. It arises due to separation of charges. $e=B \times($ rate of change of area of loop $)$
If $\theta$ is the angle between the rod and $l$ the radius of circle at $X$ at time $t,$ the area of the arc formed by the rod and radius is
Area$(O X Y)=\frac{1}{2} l^{2} \theta$
where, $l$ is radius of the circle.
$\therefore e=B \times \frac{d}{d t}\left(\frac{1}{2} l^{2} \theta\right)$
$\Rightarrow=\frac{1}{2} B \cdot l^{2} \frac{d \theta}{d t}$
$=\frac{1}{2} B l^{2} \omega \left(\because \omega=\frac{d \theta}{d t}\right)$