Magnetics 2 Question 21
21. The wire loop $P Q R S P$ formed by joining two semicircular wires of radii $R _1$ and $R _2$ carries a current $I$ as shown. The magnitude of the magnetic induction at the centre $C$ is ……
(1988, 2M)
Analytical & Descriptive Questions
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Answer:
Correct Answer: 21. $\frac{\mu _0 I}{4} \frac{1}{R _1}-\frac{1}{R _2}$ (perpendicular to paper outwards)
Solution:
- At $C$ magnetic field due to wires $P Q$ and $R S$ will be zero. Due to wire $Q R$,
$$ B _1=\frac{1}{2} \frac{\mu _0 I}{2 R _1}=\frac{\mu _0 I}{4 R _1} \quad \text { (perpendicular to paper outwards) } $$
And due to wire $S P$,
$$ B _2=\frac{1}{2} \frac{\mu _0 I}{2 R _2}=\frac{\mu _0 I}{4 R _2} \quad \text { (perpendicular to paper inwards) } $$
$\therefore$ Net magnetic field would be,
$$ B=\frac{\mu _0 I}{4} \frac{1}{R _1}-\frac{1}{R _2} \quad \text { (perpendicular to paper outwards) } $$
