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Q. The wire loop PQRSP formed by joining two semicircular wires of radii $R_1 and R_2$ carries a current 7 as shown. The magnitude of the magnetic induction at the centre C i s .......Physics Question Image

IIT JEEIIT JEE 1988Moving Charges and Magnetism

Solution:

At C magnetic field due to wires PQ and RS will be zero. Due
to wire QR
$B_1=\frac{1}{2}\Bigg(\frac{\mu_0 I}{2R_1}\Bigg)=\frac{\mu_0 I}{4R_1}$ (perpendicular to paper outwards)
And due to wire SP
$B_2=\frac{1}{2}\Bigg(\frac{\mu_0 I}{2R_2}\Bigg)=\frac{\mu_0 I}{4R_2}$ (perpendicular to paper inwards)
$\therefore $ Net magnetic field would be,
$B=\frac{\mu_0 I}{4}\Bigg(\frac{1}{R_1}-\frac{1}{R_2}\Bigg)$ (perpendicular to paper outwards)