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Question

A moving coil galvanometer has a coil of area A and number of turns N. A magnetic field B is applied on it. Moment of inertia of the coil is I about the axis of rotation.

(a) Find the torsion constant, C of the spring, if a current i0 produces a deflection of π2 in the coil.

(b) If a charge Q is passed instantaneously through the coil, find the maximum angular deflection, θ0 in it.

(Maximum torsional energy=12Cθ20)

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Solution

The correct option is **C** θ0=Q√πNAB2i0I

(a) Torque acting on the coil when angle of defection is π2,

τ=Ni0ABsin90∘=Ni0AB .....(1)

The torsion constant of a coil is,

C=τθ

C=Ni0ABπ/2=2Ni0ABπ .....(2)

So, option (B) is the correct option.

(b). If θ0 is the maximum deflection, then by conservation of energy

12Cθ20=12Iω2

⇒θ0=√IC×ω .........(3)

If L is the angular momentum of the coil, then τ=dLdt, and equation (1) can be written as

dLdt=NdQdtAB

⇒dL=NABdQ

On integrating, we get

∫L0dL=NAB∫Q0dQ

⇒L=NABQ

⇒Iω=NABQ [∵L=Iω]

⇒ω=NABQI .......(4)

Using equations (2),(3) and (4), we get

θ0= ⎷I2Ni0ABπ×NABQI

∴θ0=Q√πNAB2i0I

So, option (C) is also correct.

(a) Torque acting on the coil when angle of defection is π2,

τ=Ni0ABsin90∘=Ni0AB .....(1)

The torsion constant of a coil is,

C=τθ

C=Ni0ABπ/2=2Ni0ABπ .....(2)

So, option (B) is the correct option.

(b). If θ0 is the maximum deflection, then by conservation of energy

12Cθ20=12Iω2

⇒θ0=√IC×ω .........(3)

If L is the angular momentum of the coil, then τ=dLdt, and equation (1) can be written as

dLdt=NdQdtAB

⇒dL=NABdQ

On integrating, we get

∫L0dL=NAB∫Q0dQ

⇒L=NABQ

⇒Iω=NABQ [∵L=Iω]

⇒ω=NABQI .......(4)

Using equations (2),(3) and (4), we get

θ0= ⎷I2Ni0ABπ×NABQI

∴θ0=Q√πNAB2i0I

So, option (C) is also correct.

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