Question

# A metal wire PQ of mass 10 g lies at rest on two horizontal metal rails separated by 5 cm as shown in fig. A vertically downward magnetic field of magnitude 0.80 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below $$20.0\Omega$$, the wire PQ starts sliding on the rails. Find the coefficient of friction.

Solution

## For equilibrium, the magnetic force is equal to the friction force.$$BIl=\mu mg$$ where $$\mu$$ is the coefficient of friction and current $$I=V/R=6/20 A$$$$\Rightarrow \displaystyle \mu=\frac{BIl}{mg}=\frac{0.80\times(6/20)\times (5/100)}{(10/1000)(10)}=0.12=3/25$$  PhysicsNCERTStandard XII

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