Ncert Exercises & Solutions

A current-carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z, and z-x planes with their centers at the origin, joined together. Find the direction and magnitude of B at the origin.

Hint: Use the formula of the magnetic field due to a circular loop.

Step 1: For the current-carrying loop quarter circle of radius R, lying in the positive quadrants of the x-y plane:

B_{1} = \frac{μ_{0}}{4 π} \frac{I (π / 2)}{R} \hat{k} = \frac{μ_{0}}{4} \frac{I}{2 R} \hat{k}

For the current-carrying loop quarter circle of radius R, lying in the positive quadrants of the y-z plane:

B_{2} = \frac{μ_{0}}{4} \frac{I}{2 R} \hat{i}

For the current-carrying loop quarter circle of radius R, lying in the positive quadrants of the z-x plane:

B_{3} = \frac{μ_{0}}{4} \frac{I}{2 R} \hat{j}

Step 2: The net magnetic field is equal to the vector sum of the magnetic field due to each quarter and given by

B = \frac{1}{4 π} (\hat{i} + \hat{j} + \hat{k}) \frac{μ_{0} I}{2 R}

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