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Question

Obtain the expression for the deflecting torque acting on the current carrying rectangular coil of a galvanometer in a uniform magnetic field. Why is a radial magnetic field employed in the moving coil galvanometer ?

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Solution

Let us take a rectangular coil ABCD carrying current I and number of turns area N. A uniform magnetic field B exists in the plane of the coil along x-direction
As the side BC and DA is along the magnetic field B. So Force on BC and DA is zero due to the magnetic field.
Force will be only acting on AB and CD.
Since B is same everywhere in plane and current I is flowing in the coil. So force on both AB and CD will be equal in magnitude but diections.
If θ is angle turned by the coil.
F=(Il2B)N^K
Torque =τ=F×l12sinθ+F×l12sinθ

τ=Fl1sinθ2×2=(Fl1sinθ2)×2

τ=(BIl1l2sinθ2)×2N=NIBl1l2sinθ

Let M=NIA
Where, A=l1l2= cross - sectional area of coil
τ=NIABsinθ
τ=MBsinθ
Torque =τ=M×B (cross -product of M and B)
where, M is called as magnetic moment of the coil.
As the radial magnetic field provides constant torque on the coil irrespective of the rotation of the coil.
This makes the deflection directly proportional to the current and we can easily take the measurement of the galvanometer scale.

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