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

 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)

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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