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  • A long wire bent as shown in figure carries carries t. If the radius of the semicirc

A long wire bent as shown in figure carries carries t. If the radius of the semicircular portion is a, the magnetic field at the center C is

Option: 1

\frac{\mu_0 I}{4 a}


Option: 2

\frac{\mu_0 I}{4 \pi a} \sqrt{x^2+4}.


Option: 3

\frac{\mu_0 I}{4 a}+\frac{\mu_0 I}{4 \pi a}


Option: 4

\frac{\mu_0 I}{4 \pi a} (\sqrt{\pi^2-4})


Answers (1)

best_answer

B_1=\frac{\mu_0}{4 \pi} \times \frac{2 \pi I}{a} \times \frac{1}{2} (due to semicircular part)

 

B_2=\frac{\mu_0}{4 \pi} \times \frac{2 I}{a} (due to parallel parts of currents)

There two fields are at right angles to each others. Hence, Resultant field B =\sqrt{B_1^2+B_2^2} \\

                                         =\frac{\mu_0 I}{4 \pi a} \sqrt{\pi^2+4}

Posted by

Shailly goel

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