Question

A uniform magnetic field of induction $B$ fills a cylindrical volume of radius $R$. A rod $AB$ of length $2l$ is placed as shown in figure. If $B$ is changing at the rate $dB/dt$, the emf that is produced by the changing magnetic field and that acts between the ends of the rod is

• A. $\frac{dB}{dt}l\sqrt{R^2-l^2}$
• B. $\frac{dB}{dt}l\sqrt{R^2+l^2}$
• C. $\frac{1}{2}\frac{dB}{dt}l\sqrt{R^2-l^2}$
• D. $\frac{1}{2}\frac{dB}{dt}l\sqrt{R^2+l^2}$

Answer 1

The correct option is A $\frac{dB}{dt}l\sqrt{R^2-l^2}$

Emf induced between the ends of the rod $E=A.\frac{dB}{dt}$

where $A$ is the area of the triangular $AOB$

From figure, $OC=\sqrt{O{B}^2-B{C}^2}$

$\therefore$ $OC=\sqrt{R^2-l^2}$

Area of the triangle $A=\frac{1}{2}\times OC\times AB$

$\therefore$ $A=\frac{1}{2}\sqrt{R^2-l^2}\left(2l\right)=l\sqrt{R^2-l^2}$

$⟹$ Emf induced between the ends of the rod $E=\frac{dB}{dt}l\sqrt{R^2-l^2}$