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Question: The sensitivity of tangent galvanometer is maximum when deflection is (A) \({45^o}\) (B) \({90^o...

The sensitivity of tangent galvanometer is maximum when deflection is
(A) 45o{45^o}
(B) 90o{90^o}
(C) 30o{30^o}
(D) 60o{60^o}

Explanation

Solution

Hint The angle of deflection per unit current flowing through the galvanometer is known as the sensitivity of the galvanometer. Therefore consider an elementary current dIdI through a deflection of θ\theta and apply the tangent law.

Compete Step by step solution
According to the tangent law,
B=BHtanθB = {B_H}\tan \theta
where, BB is the magnetic field produced inside the coil due to the current II passing through it.
BH{B_H} is the horizontal component of the magnetic field.
For a current- carrying conductor,
B=nμoI2aB = \dfrac{{n{\mu _o}I}}{{2a}}
where, nn is the number of loops.
Therefore equating the above two equations we get
nμoI2a=BHtanθ\dfrac{{n{\mu _o}I}}{{2a}} = {B_H}\tan \theta
I=2aBHtanθnμo\Rightarrow I = \dfrac{{2a{B_H}\tan \theta }}{{n{\mu _o}}}
Let us consider K=BHGK = \dfrac{{{B_H}}}{G} where G=nμo2aG = \dfrac{{n{\mu _o}}}{{2a}}
I=Ktanθ\therefore I = K\tan \theta
Let us differentiate both sides.
dI=Ksec2θdθ\therefore dI = K{\sec ^2}\theta d\theta
Here dIdI stands for an elementary change in the current flowing through the galvanometer, and dθd\theta stands for an elementary change in the angle of deflection, or the angle through which the needle will turn inside the galvanometer.
Therefore we can say that dθdI\dfrac{{d\theta }}{{dI}} is the sensitivity of the galvanometer.
Dividing both sides by IIwe get,
dII=Ksec2θdθI\dfrac{{dI}}{I} = \dfrac{{K{{\sec }^2}\theta d\theta }}{I}
dII=Ksec2θdθKtanθ\Rightarrow \dfrac{{dI}}{I} = \dfrac{{K{{\sec }^2}\theta d\theta }}{{K\tan \theta }}
dII=dθsinθcosθ=2dθ2sinθcosθ\Rightarrow \dfrac{{dI}}{I} = \dfrac{{d\theta }}{{\sin \theta \cos \theta }} = \dfrac{{2d\theta }}{{2\sin \theta \cos \theta }}
We know that sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta
Substituting this value within the above equation we get,
dII=2dθsin2θ\dfrac{{dI}}{I} = \dfrac{{2d\theta }}{{\sin 2\theta }}
dθ=sin2θ2dII\Rightarrow d\theta = \dfrac{{\sin 2\theta }}{2}\dfrac{{dI}}{I}
The maximum sensitivity of the tangent galvanometer means the maximum possible value of dθd\theta .
From the above equation, we can see that dθd\theta is maximum when sin2θ\sin 2\theta is maximum.
sin2θ=1\therefore \sin 2\theta = 1
2θ=90o\Rightarrow 2\theta = {90^o}
θ=45o\Rightarrow \theta = {45^o}

Therefore, option A. is the correct answer.

Additional Information
For the measurement of small electric currents, tangent galvanometers are used. Hence a tangent galvanometer is very sensitive and accurate. For a fractional change in the value of current, there is a large deflection in the galvanometer. It consists of a coil made from an insulated copper wire wound on a circular non-magnetic frame.

Note A tangent galvanometer is dependent on the law of tangent magnetism, hence it should not be confused with a moving coil galvanometer. A moving coil galvanometer works when it is acted upon by a torque.