Question

A circular disc of mass m and radius r is rotating with an angular velocity $$\displaystyle \omega$$ 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$$\displaystyle \mu$$

A
The induced emf across the terminals of the switch is Bωr2
B
The switch is closed at t = 0 the torque required is (1/4)B2r2ω2+μmg (to maintain the constant angular speed at steady state)
C
The current in the circuit as a function of time will be given as Br2ω2R(1eRL)
D
The switch is closed at t = 0 the torque required is (1/2)B2r2ω2+(2/3)μmg ( to maintain the constant speed at steady state)

Solution

The correct option is D The switch is closed at t = 0 the torque required is (1/2)$$\displaystyle B^{2}r^{2}\omega^{2}+\left ( 2/3 \right )\mu mg$$ ( to maintain the constant speed at steady state)$$\displaystyle \epsilon =\left [ d\varepsilon =vBdr \right ]$$$$\displaystyle e=B\infty \int_{0}^{f}rdr=\frac{B\omega r^{2}}{2}$$when the Switch is closed Applying kichoff;s low $$\displaystyle -IR-L\frac{dj}{dt}+\epsilon =0$$$$\displaystyle -iR-L\frac{di}{dt}+\frac{B\omega r^{2}}{2}=0$$$$\displaystyle \frac{Br^{2}\omega -2iR}{2}=L\frac{dj}{dt};\int_{0}^{i}\frac{di}{Br^{2}-2iR}=\int_{0}^{t}\frac{dt}{2L}$$Which gives, $$\displaystyle \frac{1}{2R}\left |\epsilon n(Br^{2}\omega -2R) \right |\left | \frac{t}{2L} \right |_{0}^{t}$$Which gives i = $$\displaystyle \frac{Br^{2}\omega }{2R}(1-e^{\frac{R}{L}})$$Consider a differential circular strip on the disc of X and thicknes dx Mass of the strip $$\displaystyle dm=p2 \pi xdx,$$Wehre $$\displaystyle p=\frac{M}{\pi r^{2}}$$Frictional force on this strip along the tangent dF = $$\displaystyle \mu (dmg)=\mu 2\pi x dxg$$$$\displaystyle \mu =\ l d\mu =\mu gp2\pi(r^{3}/3)=(2/3)\mu mgr$$Torque on the strip due to friction $$\displaystyle \mu =\jmath d\mu =\mu gp2\pi(r^{3}/3)=(2/3)\mu mgr$$And also to magnetic force at steady state $$\displaystyle t_{m}=dt_{m}=\frac{Br^{2}\omega }{2R}B\int_{0}^{r}xdx=\frac{1}{4}B^{2}r^{2}\omega$$Hence torque required to maintain constant angular speed at steady state $$\displaystyle =t_{m}+\mu =(1/4)B^{2}r^{2}\omega +(2/3)\mu mg$$Physics

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