Question

A rod of length $1.5\text{m}$ is rotating about $\text{O}$ over a frictionless conducting ring in a magnetic field $B=0.4\text{T},$ with constant angular velocity $\omega =10\text{rad/s}$ as shown in the figure. Then the current flowing through the rod $\text{OA}$ is:

• A. $2.25\text{A}$
• B. $4.5\text{A}$
• C. $3.45\text{A}$
• D. $1.05\text{A}$

Answer 1

The correct option is A $2.25\text{A}$

Emf developed across the rod is,

$E=\frac{B\omega l^2}{2}$

$\therefore E=\frac{0.4\times 10\times (1.5{)}^2}{2}=4.5\text{V}$

The equivalent circuit can be drawn as,


The equivalent resistance of the circuit is,

$R_{eq}=\frac{3\times 6}{3+6}=\frac{18}{9}=2\Omega$

Thus, current flowing through the rod is,

$i=\frac{E}{R_{eq}}=\frac{4.5}{2}=2.25\text{A}$

Hence, $\left(\text{A}\right)$ is the correct answer.