Q. 23

# A metal wire PQ of mass 10 g 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.

Given-

Mass of the metal wire, M = 10 g

Distance between the two metallic strips, l = 4.90 cm

Magnetic field acting vertically-downward, B = 0.800 T

Given in the question that when the resistance of the circuit is slowly decreased below 20.0Ω the wire end PQ starts sliding on the metallic rails.

At that time, current I from ohm’s law becomes

where,

v= applied voltage and

R = resistance of the circuit

From Fleming’s left-hand rule, we can confer that the magnetic force will act towards the right.

This magnetic force will make the wire glide on the rails.

The frictional force present at the surface of the metallic rails will try to oppose this motion of the wire.

When the wire starts sliding on the rails, the frictional force acting on the wire present between the wire and the metallic rail will just balance the magnetic force acting on the wire due to current flowing through it.

Thus,

,

where

μ is the coefficient of friction

R is the normal reaction force and

F is the magnetic force

frictional force will be equal.

where

μ is the coefficient of friction for the two surfaces

W is the weight of the object

= mass × acceleration due to gravity

= mg

Also,

Magnetic force due to presence of current given by –

where,

B= magnetic field

I = current

l = length of the wire

and θ = the angle between B and l

Hence,

substituting values

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