Magnetism> Solve this problem please...................

Solve this problem please........................................

Syed suha , 7 Years ago

Grade 12

2 Answers

Khimraj

Magnetic field due to a wire of finite length is given by

B = (u_oI/4 $\pi$ r)(sin $\Theta 1$ + sin $\Theta 2$ )

where $\Theta 2$ is angle QOM

and $\Theta 1$ is angle POM

r is perpendicular distance

So B = (u_oI/4 $\pi$ a)(5a/3 + 5a/3)

So B = (5/6)u_oI/ $\pi$ a.

Hope it clears. If you like answer then please approve it.

Last Activity: 7 Years ago

Khimraj

Magnetic field due to a wire of finite length is given by

B = (u_oI/4 $\pi$ r)(sin $\Theta 1$ + sin $\Theta 2$ )

where $\Theta 2$ is angle QOM

and $\Theta 1$ is angle POM

r is perpendicular distance

So B = (u_oI/4 $\pi$ a)(3/5 + 4/5)

So B = (7/20)u_oI/ $\pi$ a.

Hope it clears. If you like answer then please approve it.

Approved

Last Activity: 7 Years ago

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Magnetism> Solve this problem please...................

Solve this problem please........................................

Syed suha , 7 Years ago

Khimraj

Khimraj

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