Question

A wire $\text{PQ}$ of mass $10\text{g}$ is at rest on two parallel metal 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 rails, directed downwards. The net resistance of the entire circuit is slowly decreased by decreasing the resistance of a variable resistor. When the resistance decreases to below $20\Omega$, the wire $\text{PQ}$ just begins to slide on rails. The coefficient of friction between wire and rails will be:

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

Answer 1

The correct option is C $0.12$

At the limiting case, when the wire $\text{PQ}$ just begins to slide, the limiting friction will act.

$f_{max}=\mu N=\mu mg..........\left(1\right)$

In this condition $R=20\Omega ,E=6\text{V}$

The direction of current in circuit will be from +ve terminal of battery to -ve terminal, thus direction of current in wire will be from $\text{P}$ to $\text{Q}$.

Hence, using $i(\overrightarrow{L}\times \overrightarrow{B})=F_m$, the magnetic force will act towards right on wire as shown in the figure below.


At sliding condition,

$f_{max}=F_m$

$\mu mg=BiL\sin {90}^{\circ }$

$\mu (10\times {10}^{-3}\times 9.8)=0.80\times \left(\frac{E}{R}\right)\times (4.9\times {10}^{-2})$

$\mu \times 10\times {10}^{-3}\times 9.8=0.80\times \left(\frac{6}{20}\right)\times 4.9\times {10}^{-2}$

$\mu =\frac{0.80\times 6\times 4.9}{20\times 9.8}$

$\mu =\frac{0.80\times 6}{40}$

$\therefore \mu =0.12$

Hence, option $\left(C\right)$ is the correct answer.