Physics Question on Electromagnetism

Question

The given circuit shows a uniform straight wire AB of 40 cm length fixed at both ends. In order to get zero reading in the galvanometer G, the free end of J is to be placed from B at:
Problem Fig

Answer 1

Solution

For zero galvanometer deflection, the bridge must be balanced. This is a Wheatstone bridge arrangement.

The condition for a balanced Wheatstone bridge is:

$\frac{R_1}{R_2} = \frac{R_3}{R_4}$

Where R1, R2, R3, and R4 represent the resistance in each arm of the Wheatstone bridge. Here, R1 and R2 are the resistances of the wire segments AJ and JB respectively, and R3 = 8 Ω and R4 = 12 Ω.

The length of AB is 40 cm, thus, $\frac{R_1}{R_2} = \frac{8}{12} = \frac{2}{3}$

$\frac{R_1}{R_2} = \frac{x}{40 - x} = \frac{2}{3}$

3x = 80 − 2x 5x = 80 x = 16 cm

Distance of J from B = 40 cm − 16 cm = 24 cm