Question

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 1

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

Answer 2

Hello Dear.

Here is the answer---


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As per as the Questions, Iron Rod is bent in the form of the Semicircle,
Thus, Length of the Iron Rod = Circumference of Semi-Circle
                                             L  = 1/2 × 2πr
                                             L = πr

⇒  Radius of the Semicircle = r
                                             = L/π

Using the Formula,

 Magnet Dipole  =  Magnetic Dipole Movement × Minimum Separation between the two Dipoles.

Magnetic Dipole Movement = M
Minimum Separation Between the Dipoles = Diameter of the Circle
                                                                        = 2 × Radius(r)
                                                                        = 2 × L/π

Thus, Magnetic Dipole = M × 2 × L/π

Therefore, Magnetic Movement = Magnetic Dipole/Length of the Rod
                                                   = (2ML/π)/L
                                                   = 2 × M/π

Thus, the Magnetic Movement between the Dipoles is 2 × M/π


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Hope it helps.

Have a nice day.