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  • A long conducting wire carrying a current I is bent at [Fig.1.128]

A long conducting wire carrying a current I is bent at 120^{\circ} [Fig.1.128]. The magentic field B at a point on the right bisector of bending angle at a distance d from the bend is (\mu_0 is the permeability of free space).

Option: 1

\frac{3 \mu_0 I}{2 \pi d}


Option: 2

\frac{\mu_0 I}{2 \pi d}


Option: 3

\frac{\mu_0 I}{\sqrt{3} \pi d}


Option: 4

\frac{\sqrt{3} \mu_0 I}{2 \pi d}


Answers (1)

P Q=\text { d, } P R=\frac{d \sqrt{3}}{2}=P S

Magnetic field at P due to the Portion QT,

 B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{\frac{d \sqrt{3}}{2}}\left(\sin 90^{\circ}+\sin 30^{\circ}\right) \\

=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{d \sqrt{3}} \cdot \frac{3}{2}=\frac{\sqrt{3} \mu_0 I}{4 \pi d}

Magnetic field at P due to the portion QU, B_2=\frac{\sqrt{3} \mu_0 I}{4 \pi d}

\thereforeTotal magnetic field at  P=B_1+B_2=\frac{\sqrt{3} \mu_0 I}{2 \pi d}

The option D is correct.

Posted by

Sumit Saini

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