A Conductor (shown in the figure) carrying constant current I is kept in the x - y plane in a uniform magnetic field → B . If F is the magnitude of the total magnetic force acting on the conductor, then the correct

Correct answer: A. If → B is along ^ z , F ∝ ( L + R ) → F = I( → Displacement ) × → B As magnetic field is uniform → F = I(2L + 2R) ^ i × → B When → B is along y or z axis F ∝ (L + R) When → B is along x -axis F = 0

JEE Advanced 2015Paper 1

Exam: JEE Advanced
Year: 2015
Paper: Paper 1
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Advanced 2015 · Paper 1 · Physics · Magnetic Effects of Current

A Conductor (shown in the figure) carrying constant current I is...

A Conductor (shown in the figure) carrying constant current $I$ is kept in the $x$ - $y$ plane in a uniform magnetic field $\overset{\to}{B}$ . If $F$ is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is (are)

Options

AIf $\overset{\to}{B}$ is along $\overset{^}{z}$ , $F ∝ ( L + R )$
BIf $\overset{\to}{B}$ is along $\overset{^}{x}$ , $F = 0$
CIf $\overset{\to}{B}$ is along $\overset{^}{y}$ , $F ∝ ( L + R )$
DIf $\overset{\to}{B}$ is along $\overset{^}{z}$ , $F = 0$

Correct answer

A. If $\overset{\to}{B}$ is along $\overset{^}{z}$ , $F ∝ ( L + R )$

Step-by-step solution

\overset{\to}{F} = I\left(\overset{\to}{Displacement}\right) \times \overset{\to}{B}

As magnetic field is uniform

\overset{\to}{F} = I\left(2L + 2R\right)\overset{^}{i} \times \overset{\to}{B}

When

\overset{\to}{B}

is along

y

z

axis

F ∝ \left(L + R\right)

When

\overset{\to}{B}

is along

x

-axis

F = 0

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