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Question

A conducting rod AB of mass M and length L is hinged at its end A. It can rotate freely in the vertical plane (in the plane of figure). A long straight wire is vertical and carrying current I. The wire passes very close to A. If the rod is released from its vertical position of unstable equilibrium, choose the correct statements at the instant shown in the figure.


A
The emf between the ends of the rod when it has rotated through an angle θ. (see figure).
E=μ0I2πsinθ3gL(1cosθ)
 
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B
The emf between the ends of the rod when it has rotated through an angle θ. (see figure)
E=μ0IπsinθgL(1cosθ)
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C
The angular velocity at that instant is,
ω=3g(1cosθ)L
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D
The angular velocity at that instant is, ω=g(1cosθ)L
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Solution

The correct options are
A The emf between the ends of the rod when it has rotated through an angle θ. (see figure).
E=μ0I2πsinθ3gL(1cosθ)
 
C The angular velocity at that instant is,
ω=3g(1cosθ)L

If we apply the energy conservation principle to find the angular speed ω of the rod,
12IAω2= loss in gravitational potential energy
12(ML23)ω2=Mgh
here h is change in position of COM of rod
h=L2[1cosθ]

12(ML23)ω2=MgL2[1cosθ]
ω=3g(1cosθ)L

Now, consider an element of length dx on the rod.
Speed of the element is, v=ωx
Magnetic field at the location of the element due to current carrying wire is,
B=μ0I2πd=μ0I2πxsinθ
EMF induced in the element is,
dE=Bvdx=μ0Iω2πsinθdx
Emf in the rod is, E=dE=μ0Iω2πsinθL0dx

E=μ0IωL2πsinθ
E=μ0I2πsinθ3gL(1cosθ)

Physics

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