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 $0.80$ tesla is applied perpendicular to the plane of the rails, directed inwards. The resistance of the circuit is slowly decreased. When the resistance decreases to below $200\Omega$, the wire $\text{PQ}$ begins to slide on the rails. The coefficient of friction between the wire and the rails is

• A. $0.12$
• B. $0.24$
• C. $0.18$
• D. $0.32$

Answer 1

The correct option is A $0.12$

At the instant, when the wire $\text{PQ}$ begins to slide, the magnetic force is just equal the force of friction.

$F_f=F_{mg}$

$\mu mg=Bil\sin \theta ........\left(1\right)$

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

Eq $\left(1\right)$ becomes

$\mu \times 10\times {10}^{-3}\times 9.8=0.80\times 0.3\times 4.9\times {10}^{-2}$

$\Rightarrow \mu =0.4\times 0.3=0.12$

Hence, option (a) is the correct answer.