question_answer

A magnet of length I, and moment M is cut into two halves (A and B) perpendicular to its axis. One piece A is bent into a semicircle of radius R and is joined to the other piece at the poles as shown in the figure below. Assuming that the magnet is in the form of a thin wire initially, the moment of the resulting moment is given by

A)  \[\frac{M}{2\pi }\]                                         

B)  \[\frac{M}{\pi }\]

C)  \[\frac{M\left( 2+\pi  \right)}{2\pi }\]                                    

D)  \[\frac{M\pi }{2+\pi }\]

Correct Answer: C

Solution :

  \[M=(m\times L)\]where m = pole strength of magnetic dipole, Magnetic dipole moment of A. \[M=m\times (2R)\]                 Since,                    \[R\times \pi =\frac{L}{2}\]                 or                            \[2R=\frac{L}{\pi }\]                 \[\therefore \]                  \[M=\left( \frac{mL}{\pi } \right)\]                 or                            \[M=\left( \frac{M}{\pi } \right)\] The magnetic moment of B \[M\,=m\times \frac{L}{2}\] \[M\,=\left( \frac{M}{2} \right)\]                 Hence, resultant magnetic moment                 \[{{M}_{R}}=M+M\,=\frac{M}{\pi }+\frac{M}{2}=M\left( \frac{2+\pi }{2\pi } \right)\]