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  • The magnetic field at O due to the wire loop A B C D carrying a current I as shown in Fig. is <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/17/17.png" style

The magnetic field at O due to the wire loop A B C D carrying a current I as shown in Fig. is

 

Option: 1

\mathrm{\frac{\mu_0 I}{4 r} }


Option: 2

\mathrm{\frac{3 \mu_0 I}{8 r} }


Option: 3

\mathrm{\frac{5 \mu_0 I}{12 r} }


Option: 4

\mathrm{\frac{7 \mu_0 I}{16 r}}


Answers (1)

best_answer

Magnetic field at O due to straight segments B C and A D is zero. For the curved part \mathrm{A B, \theta=\frac{3 \pi}{2}}  Therefore,

\mathrm{B_{A B}=\frac{3 \mu_0 I}{8 r} \quad}  directed into the page. 

For the curved part C D,\mathrm{ \theta=\frac{\pi}{2}} . Therefore,

\mathrm{B_{C D}=\frac{\mu_0 I}{16 r} \quad}    directed into the page. 
Therefore, magnetic field at O due to current I in A B C D is

\mathrm{B =B_{A B}+B_{C D} } 
\mathrm{=\frac{3 \mu_0 I}{8 r}+\frac{\mu_0 I}{16 r}=\frac{7 \mu_0 I}{16 r}}

directed into the page. So the correct choice is (d).

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Riya

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