Ncert Exercises & Solutions

A current-carrying circular loop of radius R is placed in the x-y plane with center at the origin. Half of the loop with x > 0 is now bent so that it now lies in the y-z plane.

1.	The magnitude of the magnetic moment now diminishes.
2.	The magnetic moment does not change.
3.	The magnitude of B at (0, 0, z), z >>R increases.
4.	The magnitude of B at (0, 0, z), z >>R is unchanged.

(a) Hint: Apply Fleming's left-hand rule to find the direction of the resultant magnetic field.

Step 1: Initially, the magnetic moment is given by;

M = I \times {πR}^{2}

Step 2: When the loop is bent, the magnetic moment of each half-loop;

M_{1} = M_{2} = \frac{I \times {πR}^{2}}{2}

So, the net magnetic moment is given by;

M' = \sqrt{{M_{1}}^{2} + {M_{2}}^{2}} = \frac{I \times {πR}^{2}}{2} \times \sqrt{2} = \frac{I \times {πR}^{2}}{\sqrt{2}} < M

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