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प्रश्न
Consider the situation shown in the figure. Suppose the circular loop lies in a vertical plane. The rod has a mass m. The rod and the loop have negligible resistances but the wire connecting O and C has a resistance R. The rod is made to rotate with a uniform angular velocity ω in the clockwise direction by applying a force at the midpoint of OA in a direction perpendicular to it. Find the magnitude of this force when the rod makes an angle θ with the vertical.
उत्तर
When the circular loop is in the vertical plane, it tends to rotate in the clockwise direction because of its weight.
Let the force applied be F and its direction be perpendicular to the rod.
The component of mg along F is mg sin θ.
The magnetic force is in perpendicular and opposite direction to mg sin θ.
Now,
Current in the rod:-
\[i = \frac{B a^2 \omega}{2R}\]
The force on the rod is given by
\[F_B = iBl = \frac{B^2 a^2 \omega}{2R}\]
Net force \[= F −\frac{B^2 a^2 \omega}{2R}+mg \sin \theta\]
The net force passes through the centre of mass of the rod.
Net torque on the rod about the centre O:-
\[\tau = \left( F - \frac{B^2 a^3 \omega}{2R} + mg \sin\theta \right)\frac{OA}{2}\]
Because the rod rotates with a constant angular velocity, the net torque on it is zero.
i.e. τ = 0
\[\left( F - \frac{B^2 a^3 \omega}{2R} + mg \sin\theta \right)\frac{OA}{2} = 0\]
\[\therefore F = \frac{B^2 a^3 \omega}{2R} - mg \sin\theta\]
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