Question

a magnetised wire of moment 3.14 A-$m^2$ is bent in the form of a semi- circle ; then the new magnetic moment will be :-

• A. 3.14 $A-m^2$
• B. 2 $A-m^2$
• C. 6.28 $A-m^2$
• D. None of these

Answer 1

The correct option is B 2 $A-m^2$

$\text{Given}$ : Magnetized wire of moment = 3.14 A/$m^2$

$\text{To find}$ : New magnetic moment

$\text{Solution}$ :

Let us assume the pole strength of wire is p and the length is L, then

P$\times L$ = 3.14 A/$m^2$

The wire is then bent in the form of a semicircle

Now we know, L = 3.14 R

So, R = $\frac{L}{3.14}$

So the semicircle’s effective length = 2 R = $\frac{2L}{3.14}$

So the new magnetic moment formed in wire = P$\times 2R=\frac{2L}{3.14}\times P$ = $\frac{2}{3.14}\times 3.14$ = 2

The magnetic moment is 2 A/$m^2$

$\text{Hence B is the correct option}$