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प्रश्न
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. A battery of emf ε and a variable resistance R are connected between O and C. Neglect the resistance of the connecting wires. Let θ be the angle made by the rod from the horizontal position (show in the figure), measured in the clockwise direction. During the part of the motion 0 < θ < π/4 the only forces acting on the rod are gravity and the forces exerted by the magnetic field and the pivot. However, during the part of the motion, the resistance R is varied in such a way that the rod continues to rotate with a constant angular velocity ω. Find the value of R in terms of the given quantities.
उत्तर
It is given that the rod is rotated with angular speed in clockwise direction.
The emf induced in the rod (e) is `(Bomegaa^2)/2,` with O at the lower potential and A at the higher potential.
The equivalent circuit can be drawn as:-
\[i = \frac{e + E}{R} = \frac{\frac{1}{2}B\omega a^2 + E}{R}\]
\[ = \frac{B\omega a^2 + 2E}{2R}\]
Because the rod rotates with uniform angular velocity, the net torque about point O is zero.
Now,
Net force on the rod, Fnet = mg cos θ - ilB
Net torque, τ = (mg cos θ - ilB).(r/2) = 0
∴ mg cos θ = ilB
\[R = \frac{(B\omega a^2 + 2E)}{2R}(a \times B)\]
\[ \Rightarrow R = \frac{(B\omega a^2 + 2E)aB}{2mg \cos \theta}\]
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