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  • What will be the magnetic field due to this given current-carrying straight wire at point  P as shown in the below figure. <img alt="" src="https://cdn.entrance360.com/media/upl

What will be the magnetic field due to this given current-carrying straight wire at point  P as shown in the below figure.

 

Option: 1

\frac{\mu _0i}{9\pi r}


Option: 2

\frac{\mu _0I}{2 r}


Option: 3

0


Option: 4

\frac{\mu _0i}{4\pi r}[\cos \alpha -\cos \beta]


Answers (1)

best_answer

As we have learned

If \alpha = (90 - \phi _1 ), \beta = (90 +\phi _2)

B=\frac{\mu_{o}}{4\pi}\:\frac{i}{r}\:(\cos \alpha -\cos \beta)

-

 

 

P is at perpendicular distance r from wire 

B= \frac{\mu _0i}{4\pi r}[ \sin \phi _1+\sin \phi_2]

from figure \alpha = (90 - \phi _1 ), \beta = (90 +\phi _2)

so B = \frac{\mu _0i}{4\pi r}[\cos \alpha - \cos \beta ]

 

Posted by

Devendra Khairwa

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