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Karnataka Board PUCPUC Science Class 11

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

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Question

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.

Sum
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Solution

Given:-

Mass of the metal wire, M = 10 g

Distance between the two horizontal metal rails, l = 4.90 cm

Vertically-downward magnetic field, B = 0.800 T

As per the question, when the resistance of the circuit is slowly decreased below 20.0 Ω, the wire PQ starts sliding on the rails. At that moment,

current in the wire, i = `V/R =6/20A  `Using Fleming's left-hand rule, the magnetic force will act towards the right. So, due to this magnetic force, the wire will try to slide on the rails.The frictional force will try to oppose this motion of the wire.

When the wire just starts sliding on the rails, the frictional force acting on the wire will just balance the magnetic force acting on the wire.This implies

µR = F, where

µ is the coffiecent of friction

R is the normal reaction force and

F is the magnetic force

`⇒mu xx M xx g = ilB`

`muxx10xx10^-3xx9.8 = 6/20xx 4.9 xx 10^-2xx0.8`

`mu = (6xx4.9xx10^-2xx0.8)/(20xx10xx10^-3xx9.8)`

µ = 0.12

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Force on a Moving Charge in Uniform Magnetic and Electric Fields
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Chapter 34: Magnetic Field - Exercises [Page 232]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 34 Magnetic Field
Exercises | Q 23 | Page 232
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