JEE MAIN - Physics (2009 - No. 10)
A current loop $$ABCD$$ is held fixed on the plane of the paper as shown in the figure. The arcs $$BC$$ (radius $$= b$$) and $$DA$$ (radius $$=a$$) of the loop are joined by two straight wires $$AB$$ and $$CD$$. A steady current $$I$$ is flowing in the loop. Angle made by $$AB$$ and $$CD$$ at the origin $$O$$ is $${30^ \circ }.$$ Another straight thin wire steady current $${I_1}$$ flowing out of the plane of the paper is kept at the origin.
Due to the presence of the current $${I_1}$$ at the origin:
The forces on $$AD$$ are $$BC$$ are zero.
The magnitude of the net force on the loop is given by $${{{I_1}I} \over {4\pi }}{\mu _0}\left[ {2\left( {b - a} \right) + {\raise0.5ex\hbox{$\scriptstyle \pi $}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 3$}}\left( {a + b} \right)} \right].$$
The magnitude of the net force on the loop is given by $${{{\mu _0}I{I_1}} \over {24ab}}\left( {b - a} \right).$$
The forces on $$AB$$ and $$DC$$ are zero.
Explanation
KEY CONCEPT : $$\overrightarrow F = I\left( {\overrightarrow \ell \times \overrightarrow B } \right)$$
The force on $$AD$$ and $$BC$$ due to current $${I_1}$$ is zero. This is because the directions of current element $$I\overrightarrow {d\ell } $$ and magnetic field $$\overrightarrow B $$ are parallel.
The force on $$AD$$ and $$BC$$ due to current $${I_1}$$ is zero. This is because the directions of current element $$I\overrightarrow {d\ell } $$ and magnetic field $$\overrightarrow B $$ are parallel.
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