Question

A wire $\text{PQ}$ of mass $10\text{g}$ is at rest on two parallel metallic rails. The separation between the rails is $4.9\text{cm}$. A magnetic field of $0.80\text{T}$ is applied perpendicular to the plane of the rails, directed downwards. The resistance of the circuit is slowly decreased. When the resistance decreases to below $20$ $\Omega$, the wire $\text{PQ}$ begins to slide on the rails. Calculate the coefficient of friction between the wire and the rails.

• A. $0.24$
• B. $0.36$
• C. $0.12$
• D. $0.48$

Answer 1

The correct option is C $0.12$

Wire $\text{PQ}$ just begins to slide when magnetic force is equal to the force of friction (limiting friction).

In the diagram, we can see that magnetic force will be towards right (of the student) as
${\overrightarrow{F}}_B=(i\overrightarrow{l}\times \overrightarrow{B})$

$=i\left[l\right(-\hat{j})\times B_0(-\hat{k}\left)\right]$

$=il{B}_0\hat{i}$

$\therefore$ Force of friction will be towards left (of the student).

$\Rightarrow \mu mg=ilB$

$\Rightarrow \mu =\frac{ilB}{mg}$

Here,

$i=\frac{E}{R}=\frac{6}{20}=0.3\text{A}$

$l=4.9\text{cm}=4.9\times {10}^{-2}\text{m}$

$B=0.8\text{T}$

$\Rightarrow \mu =\frac{0.3\times 4.9\times {10}^{-2}\times 0.8}{10\times {10}^{-3}\times 9.8}$

$\therefore \mu =0.12$
​​​​​​
Hence, option (c) is the correct answer.