Question

The circular coil of a glavanometer has fifty turns. The coil has a redius of cm. The coil is placed in a radial magnetic field of 0.010 t. The torsion of the spring on which the coil hangs os 3103nmrad1 determine the deviation of the galvanometer hand if the current through the coil is 1.0 ma.

Answer 1

If the torsion constant is given as per degree, why did we convert the 60 degree to radian.

Answer 2

The deviation or deflection of the galvanometer hand is 0.05062 rad/mA.

Given:

The number of turns of the coil, n = 50.

The radius of the circular coil, r = 10 cm.

The radial magnetic field, B = 0.010 T.

The torsional constant of the spring, k = 3103 Nm/rad.

The current through the coil, I = 1.0 mA.

To Find:

We have to find the deviation of the galvanometer hand using the given details.

Solution:

The deviation (deflection/twist) of the galvanometer hand is the value indicated on the galvanometer scale by a pointer which is connected to the suspension wire. It is represented by the symbol θ.

The expression for the deviation of the galvanometer hand is given by,

θ $= \frac{nIAB}{k}$,

where A is the area of the coil.

Since it is a circular coil, the equation for its area is given by,

The area of circular coil, $A = \pi r^2$,

where r is the radius of the circular coil.

On substituting the value of r in the above equation, it becomes,

$A = \pi (10)^2 = 314.16 cm^2$.

∴, On substituting the given values in the expression for θ, we get,

θ $= \frac{(50). (1.0). (314.16). (0.010)}{3103} = \frac{157.08}{3103} = 0.05062$ rad/mA.

Hence, the deviation or deflection of the galvanometer hand is 0.05062 rad/mA.

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