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  •  The magnetic field at the origin due to the current flowing in the  wire as shown in figure below is           

 The magnetic field at the origin due to the current flowing in the  wire as shown in figure below is           

 

Option: 1

\frac{-\mu _0 I }{8 \pi a } (\hat i + \hat k )


Option: 2

\frac{\mu _0 I }{2 \pi a } (\hat i + \hat k )


Option: 3

\frac{\mu _0 I }{8 \pi a } (-\hat i + \hat k )


Option: 4

\frac{\mu _0 I }{8 \pi a } (\hat i -\hat k )


Answers (1)

best_answer

Magnetic Field Due to a Straight Wire -

B=\frac{\mu_{o}i}{4\pi r}(\sin\phi_{1}+\sin\phi_{2})

OR

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

 

Here, the wire along the x-axis and the wire in the x-z plane pass through the origin.

So the magnetic field at the origin is zero.

Now for the wire parallel to the y-axis

\phi _2=0 \ \ and \ \ \phi _1=90^0

So

B_{ 0 } = \frac{\mu _0 I }{4 \pi a\sqrt{2} }

and its direction is along

\hat {B_o}=\hat {dl} \times \hat {r} =\hat {j} \times \frac{\left ( -\hat {i} + -\hat {k}\right )}{\sqrt{2}}=\cos 45 \degree \hat (-i) +\cos 45 \degree \hat k

So

\\\\ B_{ 0 } = \frac{\mu _0 I }{4 \pi a\sqrt{2} } [\cos 45 \degree \hat (-i) +\cos 45 \degree \hat k )]=\frac{\mu _0 I }{8 \pi a } (-\hat i + \hat k )

Posted by

Kshitij

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