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Question
A metallic rod of length ‘l’ is rotated with a frequency v with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtained the expression for it.
Solution
Suppose the length of the rod is greater than the radius of the circle and rod rotates anticlockwise and suppose the direction of electrons in the rod at any instant be along +y-direction.
Suppose the direction of the magnetic field is along +z-direction.
Then, using Lorentz law, we get the following:
`vecF = -e (vecv xx vecB)`
`=> vecF = -e(vhatj xx Bhatk)`
`=> hatF = -evBhati`
Thus, the direction of force on the electrons is along −x axis.
Thus, the electrons will move towards the center i.e., the fixed end of the rod. This movement of electrons will result in current and hence it will produce emf in the rod between the fixed end and the point touching the ring.
Let θ be the angle between the rod and radius of the circle at any time t.
Then, area swept by the rod inside the circle `= 1/2 πr^2θ`
Inducced emf ` = B xx d/dt (1/2πr^2θ) =1/2πr^2B (dθ)/dt = 1/2πr^2Bω = 1/2πr^2B (2πv) = π^2r^2Bv`
NOTE: There will be an induced emf between the two ends of the rods also.
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Figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.
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