question_answer

A wooden disc of mass M and radius R has a single loop of wire wound on its circumference. It is mounted on a massless rod of length d. The ends of the rod are supported at its ends so that the rod is horizontal and disc is vertical. A uniform magnetic field \[{{B}_{0}}\] exists in vertically upward direction. When a current I is given to the wire one end of the rod leaves the support. Find least value of I.  

A) \[\frac{2Mgd}{\pi {{R}^{2}}{{B}_{0}}}\]        

B) \[\frac{2\pi Mgd}{{{R}^{2}}{{B}_{0}}}\]

C) \[\frac{Mgd}{2\pi {{R}^{2}}{{B}_{0}}}\]        

D) \[\frac{\pi Mgd}{2{{R}^{2}}{{B}_{0}}}\]

Correct Answer: C

Solution :

[c] Magnetic dipole moment of loop is \[\mu =I\pi {{R}^{2}}\] (direction as shown). Magnetic torque on the loop is \[\tau =\mu {{B}_{0}}\sin 90{}^\circ =I\pi {{R}^{2}}{{B}_{0}}\] (in the direction shown) Balancing torque on the disc gives \[\tau +{{N}_{2}}\frac{d}{2}={{N}_{1}}\frac{d}{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,{{N}_{1}}-{{N}_{2}}=\frac{2I\pi {{R}^{2}}{{B}_{0}}}{d}\]   ?.(i) Also        \[{{N}_{1}}+{{N}_{2}}=Mg\]              ....(ii) Solving (i) and (ii) gives \[{{N}_{1}}=\frac{Mg}{2}+\frac{I\pi {{R}^{2}}{{B}_{0}}}{d}\] \[{{N}_{2}}=\frac{Mg}{2}-\frac{I\pi {{R}^{2}}{{B}_{0}}}{d}\] \[{{N}_{2}}<{{N}_{1}}\] \[{{N}_{2}}\] will become zero if \[\frac{I\pi {{R}^{2}}{{B}_{0}}}{d}=\frac{Mg}{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,I=\frac{Mgd}{2\pi {{R}^{2}}{{B}_{0}}}\]