A wire $P Q$ of mass $10 g$ is at rest on two parallel metal rails. The separation between the rails is $4.9 c m$ . A magnetic field of $0.80$ tesla 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 $20 o h m$ , the wire $P Q$ begins to slide on the rails. The coefficient of friction between the wire and the rails is found to be $\frac{α β}{100}$ , where $α β$ being two digit number, $α$ and $β$ are single digit integers. Then find the value of $α + β$
$(g = 9.8 m / s^{2})$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Wire PQ begins to slide when magnetic force is just equal to the force of friction
i.e., $μ m g = i l B \sin θ (θ = 90^{0})$
Here, $i = \frac{E}{R} = \frac{6}{20} = 0.3 A$

$∴ μ = \frac{i l B}{m g} = \frac{(0.3) (4.9 \times 10^{- 2}) (0.8)}{(10 \times 10^{- 3}) (9.8)} = 0.12 \Rightarrow \frac{α β}{100} = α = 1 a n d β = 2 S o, α + β = 3$