Question

A uniform rod $PQ$ of length $l$ and mass $m$ is placed on a smooth horizontal surface. It is hinged at one of its ends and touches the conducting circular loop at another end. The system is subjected to vertical magnetic field B₀. Rod is given angular velocity ${\omega }_0$ about the axis passing through $P$ . If the time after which its angular velocity becomes half is $\frac{a}{b}\frac{mR\text{l}n2}{B^2{l}^2}$ . Find the value of $\left(a+b\right)$ , where $a$ and $b$ are smallest positive integers.

Answer 1

Answer: 7

$ϵ=\frac{1}{2}{B}_0\omega l^2$
$I=\frac{1}{2}\frac{B_0\omega l^2}{R}$
$\tau =\frac{I{l}^2{B}_0}{4}=\frac{1}{4}\frac{B_0\omega l^2}{R}{l}^2{B}_0$
$\frac{B_0^2\omega l}{4R}=\frac{m{l}^2}{3}\frac{d\omega }{dt}$
$\frac{3B^2{l}^2}{4mR}t=-\int \frac{d\omega }{dt}$
$t=\frac{4mRln2}{3B^2{l}^2}$
$$\begin{gathered} \therefore \text{a}=4\text{and b}=3 \\ \Rightarrow \text{a}+\text{b}=4+3=7 \end{gathered}$$