A metal wire PQ of mass 10 gm lies at rest on two horizontal metal rails separated by 4.90 cm (figure). A vertically downward magnetic field of magnitude 0.800 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20.0 Ω, the wire PQ starts sliding on the rails. Find the coefficient of friction. Neglect magnetic force acting on wire P Q due to metal rails (g=9.8 m / s 2)

Question

A metal wire PQ of mass 10 gm lies at rest on two horizontal metal rails separated by 4.90 cm (figure). A vertically downward magnetic field of magnitude 0.800 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20.0 Ω, the wire PQ starts sliding on the rails. Find the coefficient of friction. Neglect magnetic force acting on wire P Q due to metal rails (g=9.8 m / s 2) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

F = f L N = mg BiL =μ mg μ =( BIL / mg ) =(0.8 × (6/20) × 4.9 × 10-2/10 × 10-3 × 9.8) = 0.12

F=fL​ N=mg BiL=μmg μ=mgBIL​ =10×10−3×9.80.8×206​×4.9×10−2​ =0.12

$F = f_{L}$
$N = m g$
$B i L = μ m g$
$μ = \frac{B I L}{m g}$
$= \frac{0.8 \times \frac{6}{20} \times 4.9 \times 1 0 ^{- 2}}{10 \times 1 0 ^{- 3} \times 9.8}$
$= 0.12$