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  • A current I flow in an infinetly long wire cross - section in form of a semicircular ring of radius R. The magnitude of the magnetic field induction along its axis is<st

A current I flow in an infinetly long wire cross - section in form of a semicircular ring of radius R. The magnitude of the magnetic field induction along its axis is

Option: 1

\frac{\mu oI}{2\pi ^{2}R}


Option: 2

\frac{\mu oI}{2\pi R}


Option: 3

\frac{\mu oI}{4\pi R}


Option: 4

\frac{\mu oI}{\pi ^{2}R}


Answers (1)

best_answer

As we learned

Magnetic field due to Current Element -

\overrightarrow{dB}=K \tfrac{idl\sin \Theta }{r^{2}} \Rightarrow \overrightarrow{B} =\int \overrightarrow{dB} = \frac{\mu _{0}}{4\pi } \int \frac{dl\sin \Theta }{^{r^{2}}}

-

 dI = \frac{d\Theta }{\pi}I

dB=\frac{\mu o}{4\pi}\cdot \frac{2I}{R}=\frac{\mu oI}{2\pi^{2}R}d\Theta \rightarrow M.Fat centre due to this portion

Net magnetic field

B=\int_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}dBcos\Theta \rightarrow  

B=\int_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}\frac{\mu oI}{2\pi^{2}R}cos\Theta d\Theta = \frac{\mu oI}{\pi^{2}R}

 

Posted by

Divya Prakash Singh

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