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:

Question

A charged particle moving with a velocity $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$ in a magnetic field $\vec{B}$ experiences a force $\vec{F} = F_1\hat{i} + F_2\hat{j}$. Here $v_1, v_2, F_1, F_2$ are all constants. Then $\vec{B}$ can be:

Answer 1

A balanced Wheatstone bridge arrangement yields zero galvanometer deflection. The condition for this balance is:

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

Here, R₁, R₂, R₃, and R₄ denote the resistance in each arm. Specifically, R₁ and R₂ are the resistances of wire segments AJ and JB. R₃ is 8 Ω and R₄ is 12 Ω.

Given the length of AB is 40 cm, the resistance ratio is $\frac{R_1}{R_2} = \frac{8}{12} = \frac{2}{3}$

The ratio of resistances is also expressed as $\frac{R_1}{R_2} = \frac{x}{40 - x}$, where x is the length of AJ. Thus, $\frac{x}{40 - x} = \frac{2}{3}$

Solving for x: 3x = 80 − 2x. This simplifies to 5x = 80, yielding x = 16 cm.

The distance of J from B is calculated as 40 cm − 16 cm = 24 cm.

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:

The Correct Option is D

Solution and Explanation

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