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A current I flows in an infinitely long wire with cross-section in the form of a semicircular ring of radius R The magnitude of the magnetic induction along its axis is

  • Option 1)

    \frac{\mu _{0}I}{\pi ^{2}R}

  • Option 2)

    \frac{\mu _{0}I}{2\pi ^{2}R}

  • Option 3)

    \frac{\mu _{0}I}{2\pi R}

  • Option 4)

    \frac{\mu _{0}I}{4\pi R}

 

Answers (1)

best_answer

As we learnt in 

For Finite Length -

\phi_{1}=\phi_{2}=\phi

B=\frac{\mu_{o}}{4\pi}\:\frac{i}{r}\:(2\sin\phi)

- wherein

 

 Current in a small element dI

dI = \frac{d\Theta }{\pi } I

Magnetic field due to the element

dB = \frac{\mu_0}{4\pi}.2\frac{dI}{R}

dB_{net}= dB Sin\Theta

B_{net}= \int dB Sin\Theta =\frac{\mu_I}{2\pi ^2R}\int_{0}^{\pi}sin\Theta d\Theta

=\frac{\mu_I}{\pi ^2R}


Option 1)

\frac{\mu _{0}I}{\pi ^{2}R}

Correct

Option 2)

\frac{\mu _{0}I}{2\pi ^{2}R}

Incorrect

Option 3)

\frac{\mu _{0}I}{2\pi R}

Incorrect

Option 4)

\frac{\mu _{0}I}{4\pi R}

Incorrect

Posted by

prateek

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