1.

N iron rod of length L and magnetic moment M is bent in the form of a semicircle. Now its magnetic moment will be [CPMT 1984; MP Board 1986; NCERT 1975; MP PET/PMT 1988; EAMCET (Med.) 1995; Manipal MEE 1995;RPMT 1996; BHU 1995; MP PET 2002]A) M B) \frac{2M}{\pi } C) \frac{M}{\pi } D) M\pi

Answer»

See attachment, 
length L ,long wire is bent in the shape of semicircle of RADIUS r .
then, length of wire = circumference of semicircle 
L = πr => r = L/2π 

now, MAGNETIC dipole = magnetic dipole moment × minimum separation between TWO dipoles 

here ( in PIC ) it is clear that , minimum separation between dipoles is 2r = 2L/π 

∴ magnetic dipole = M × 2L/π = 2ML/π
hence, magnetic moment =2M/π


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