A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X -axis while semicircular portion of radius R is lying in Y - Z plane. Magnetic field at

Correct answer: C. → B = - μ _ 0 4π (I)/(R)(π ^ i + 2 ^ k ) → B _ c = → B _ 1 + → B _ 2 + → B _ 3 . B _ 1 = μ _ 0 4 ( 90 + sinθ) - ^ k B _ 1 = μ _ 0 4 (- ^ k ) = B _ 3 B due to segment 2 B _ 2 = μ _ 0 I 4π × π ( - ^ i ) = - μ _ 0 I 4R ^ i So B at center → B _ c = → B _ 1 + → B _ 2 + → B _ 3 B _ c = - μ _ 0 I 4R (i + 2 ^ k π ) = - μ _ 0 I 4 ( π ^ i + 2 ^ k )

NEET 2015

Exam: NEET
Year: 2015
Subject: Physics
Chapter: Magnetic Effects of Current

NEET 2015 · Physics · Magnetic Effects of Current

A wire carrying current I has the shape as shown in adjoining...

A wire carrying current $\text{I}$ has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to $\text{X}$ -axis while semicircular portion of radius $\text{R}$ is lying in $\text{Y}$ - $\text{Z}$ plane. Magnetic field at point $\text{O}$ is :

A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and

Options

A $\overset{\to}{B} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{R}\left(\pi\overset{^}{i} + 2\overset{^}{k}\right)$
B $\overset{\to}{B} = - \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{R}\left(\pi\overset{^}{i} - 2\overset{^}{k}\right)$
C $\overset{\to}{B} = - \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{R}\left(\pi\overset{^}{i} + 2\overset{^}{k}\right)$
D $\overset{\to}{B} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{R}\left(\pi\overset{^}{i} - 2\overset{^}{k}\right)$

Correct answer

C. $\overset{\to}{B} = - \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{R}\left(\pi\overset{^}{i} + 2\overset{^}{k}\right)$

Step-by-step solution

A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and

{\overset{\to}{B}}_{c} = {\overset{\to}{B}}_{1} + {\overset{\to}{B}}_{2} + {\overset{\to}{B}}_{3} .

{B}_{1} = \dfrac{{\mu}_{0}}{4\piR}\left(\sin 90 + sin\theta\right) - \overset{^}{k}

{B}_{1} = \dfrac{{\mu}_{0}}{4\piR}\left(- \overset{^}{k}\right) = {B}_{3}

B due to segment 2

{B}_{2} = \dfrac{{\mu}_{0}I}{4\pi} \times \pi ( - \overset{^}{i} ) = - \dfrac{{\mu}_{0}I}{4R}\overset{^}{i}

So B at center

{\overset{\to}{B}}_{c} = {\overset{\to}{B}}_{1} + {\overset{\to}{B}}_{2} + {\overset{\to}{B}}_{3}

{B}_{c} = \dfrac{- {\mu}_{0}I}{4R}\left(i + \dfrac{2\overset{^}{k}}{\pi}\right) = \dfrac{- {\mu}_{0}I}{4\piR} ( \pi\overset{^}{i} + 2\overset{^}{k} )

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