Question

A circular disc of mass m and radius r is rotating with an angular velocity $\omega$ on a rough horizontal plane A uniform and constant magnetic field B is applied perpendicular and into the plane An inductor L and and external resistance R are connected through a switch S between centre of the disc O and point P The point P always touches the circumference of the disc Initially the switch S is open Take coefficient of friction between the plane and disc as$\mu$

• A. The induced emf across the terminals of the switch is $B\omega r^2$
• B. The switch is closed at t = 0 the torque required is $\left(1/4\right)B^2{r}^2{\omega }^2+\mu mg$ (to maintain the constant angular speed at steady state)
• C. The current in the circuit as a function of time will be given as $\frac{B{r}^2\omega }{2R}(1-e^{-\frac{R}{L}})$
• D. The switch is closed at t = 0 the torque required is (1/2)$B^2{r}^2{\omega }^2+\left(2/3\right)\mu mg$ ( to maintain the constant speed at steady state)

Answer 1

The correct option is D The switch is closed at t = 0 the torque required is (1/2)$B^2{r}^2{\omega }^2+\left(2/3\right)\mu mg$ ( to maintain the constant speed at steady state)

$ϵ=[d\epsilon =vBdr]$
$e=B\infty {\int }_0^{f}rdr=\frac{B\omega r^2}{2}$
when the Switch is closed Applying kichoff;s low
$-IR-L\frac{dj}{dt}+ϵ=0$
$-iR-L\frac{di}{dt}+\frac{B\omega r^2}{2}=0$
$\frac{B{r}^2\omega -2iR}{2}=L\frac{dj}{dt};{\int }_0^i\frac{di}{B{r}^2-2iR}={\int }_0^t\frac{dt}{2L}$
Which gives, $\frac{1}{2R}\left|ϵn\right(B{r}^2\omega -2R\left)\right|{\left|\frac{t}{2L}\right|}_0^t$
Which gives i = $\frac{B{r}^2\omega }{2R}(1-e^{\frac{R}{L}})$
Consider a differential circular strip on the disc of X and thicknes dx Mass of the strip $dm=p2\pi xdx,$
Wehre $p=\frac{M}{\pi r^2}$
Frictional force on this strip along the tangent dF = $\mu \left(dmg\right)=\mu 2\pi xdxg$
$\mu =ld\mu =\mu gp2\pi \left(r^3/3\right)=\left(2/3\right)\mu mgr$
Torque on the strip due to friction $\mu =ȷd\mu =\mu gp2\pi \left(r^3/3\right)=\left(2/3\right)\mu mgr$
And also to magnetic force at steady state $t_m=d{t}_m=\frac{B{r}^2\omega }{2R}B{\int }_0^{r}xdx=\frac{1}{4}{B}^2{r}^2\omega$
Hence torque required to maintain constant angular speed at steady state
$=t_m+\mu =\left(1/4\right)B^2{r}^2\omega +\left(2/3\right)\mu mg$