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Moving Charges and Magnetism — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsMoving Charges and Magnetism3 Marks
Question
Write the expression for the magnetic moment ($\overrightarrow{\text{m}}$) due to a planar square loop of side ‘l’ carrying a steady current I in a vector form. In the given figure this loop is placed in a horizontal plane near a long straight conductor carrying a steady current $I_1$ at a distance l as shown. Give reasons to explain that the loop will experience a net force but no torque. Write the expression for this force acting on the loop.
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Answer

$\overrightarrow{\text{m}} = \text{I}\overrightarrow{\text{A }} ( \overrightarrow{\text{A}} = \text{ area vector) }$
Torque: The magnetic field due to the long current carrying wire is perpendicular to the plane of paper. Hence the force acting on each of the four sides is in the plane of the paper and the net torque is zero.
Alternate Answer
$\overrightarrow{\text{m}}$is perpendicular to the plane of paper and$\overrightarrow{\text{B}}$ is perpendicular to the plane of paper. Hence $\overrightarrow{\tau} =\overrightarrow{\text{m}}\times\overrightarrow{\text{B}} = 0 $
Force:
Force on upper horizontal side $ = \frac{\text{I}l\mu_{o}\text{I}_{1}}{2\pi\text{l}} =\frac{\text{I}\mu_{o}\text{l}_{1}}{2\pi}\text{ (attractive) }$
Force on lower horizontal side $ = \frac{\text{I}l\mu_{o}\text{I}_{1}}{2\pi(2l)} = \frac{\text{I}\mu_{o}\text{I}_{1}}{4\pi}\text{ (repulsive)}$
The direction of these forces being opposite to each other therefore net force $ = \frac{\mu_{o}\text{I}_{1}\text{I}}{4\pi}\text{ (attractive) }$
(The net force on the two vertical sides is zero)
Torque: The magnetic field due to the long current carrying wire is perpendicular to the plane of paper. Hence the force acting on each of the four sides is in the plane of the paper and the net torque is zero.

Alternate Answer
$\overrightarrow{\text{m}}$is perpendicular to the plane of paper and$\overrightarrow{\text{B}}$ is perpendicular to the plane of paper. Hence $\overrightarrow{\tau} =\overrightarrow{\text{m}}\times\overrightarrow{\text{B}} = 0 $
Force: Award this mark irrespective of result obtained or calculation done by the students.

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