Two infinitely long straight wires lie in the xy - plane along the lines x = ± R . The wire located at x = + R carries a constant current I _ 1 and the wire located at x = - R carries a constant current I _ 2 . A

Correct answer: A. If I _ 1 = I _ 2 , then → B cannot be equal to zero at the origin (0, 0, 0) ( A ) At origin, → B = 0 due to two wires if I _ 1 = I _ 2 , hence( → B _ net ) at origin is equal to → B due to ring, which is non-zero. ( B ) If I _ 1 > 0and I _ 2

JEE Advanced 2018Paper 1

Exam: JEE Advanced
Year: 2018
Paper: Paper 1
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Advanced 2018 · Paper 1 · Physics · Magnetic Effects of Current

Two infinitely long straight wires lie in the xy - plane along the...

Two infinitely long straight wires lie in the xy - plane along the lines

x = \pm R

. The wire located at

x = + R

carries a constant current

{I}_{1}

and the wire located at

x = - R

carries a constant current

{I}_{2}

. A circular loop of radius R is suspended with its centre at

\left(0, 0\sqrt{3}R\right)

and in a plane parallel to the xy - plane. This loop carries a constant current

I

in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the

+ \overset{^}{j}

direction. Which of the following statements regarding the magnetic field

\overset{\to}{B}

is (are) true? Two infinitely long straight wires lie in the xy - plane along the lines x = ± R . The wire located at x = + R carries

Options

AIf ${I}_{1} = {I}_{2}, then\overset{\to}{B}$ cannot be equal to zero at the origin $\left(0, 0, 0\right)$
BIf ${I}_{1} > 0and{I}_{2} < 0, then\overset{\to}{B}$ can be equal to zero at the origin $\left(0, 0, 0\right)$
CIf ${I}_{1} < 0and{I}_{2} > 0, then\overset{\to}{B}$ can be equal to zero at the origin $\left(0, 0, 0\right)$
DIf ${I}_{1} = {I}_{2}$ , Then the z-component of the magnetic field at the centre of the loop is ((-\mu 0 $\mathrm{I}$ / 2 \mathrm{R}))

Correct answer

A. If ${I}_{1} = {I}_{2}, then\overset{\to}{B}$ cannot be equal to zero at the origin $\left(0, 0, 0\right)$

Step-by-step solution

( A )

At origin,

\overset{\to}{B} = 0

due to two wires if

{I}_{1} = {I}_{2}, hence\left({\overset{\to}{B}}_{net}\right)

at origin is equal to

\overset{\to}{B}

due to ring, which is non-zero.

( B )

{I}_{1} > 0and{I}_{2} < 0, \overset{\to}{B}

at origin due to wires will be along

+ \overset{^}{k}

direction and

\overset{\to}{B}

due to ring is along

- \overset{^}{k}

direction and hence

\overset{\to}{B}

can be zero at origin

( C )

{I}_{1} 0, \overset{\to}{B}

at origin due to wires is along

- \overset{^}{k}

and also along

- \overset{^}{k}

due to ring, hence

\overset{\to}{B}

cannot be zero

( D )

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