Chapter 20, Problem 53P

To determine

The current required to give full scale deflection in galvanometer.

Answer to Problem 53P

Solution:

Current,

$I=69.73\mu A$

Explanation of Solution

Given:

that initially galvanometer gives a full-scale deflection by 0.53µA current in its coil at a magnetic field. field drops to 0.76of its initial value.

Formula used:

Angular deflection of a galvanometer is given by

$\varphi \text{ = }\frac{\text{N}IAB}{k}$ ……….. (eq.1)

Where,

N → No of turns

I → Current through Coil

A → Area of loop

B → Field strength of magnet

k → spring constant of galvanometer

Calculation:

Given that initially galvanometer gives a full-scale deflection by 0.53µA current in its coil at a magnetic field. Equation-1 for this case can be written as

$\varphi \text{ = }\frac{\text{N}AB\times \text{53}\times {\text{10}}^{-6}}{k}$ ……….. (eq.2)

Now the field drops to 0.76 of its initial value. Let I be the current required for full scale deflection in this case. Then equation for this can be written as

$\varphi \text{ = }\frac{\text{N}AI\times 0.76\times B}{k}$ ……….. (eq.3)

Dividing eqn.3 by eqn.2 gives

$1=\frac{53\times {10}^{-6}}{I\times 0.76}$

Therefore, $I=69.73\mu A$