Two thin long parallel wires seperated by a distance 'b' are carrying a current ' I' amp each . The magnitude of the force3 per unit length exerted by one wire on the other is
A
` (mu_(0)i^(2))/(b^(2))`
B
` (mu_(0)i^(2))/(2 pib)`
C
` (mu_(0)i^(2))/( 2 pib)`
D
` (mu_(0)i^(2))/(2 pi b^(2))`
Text Solution
Verified by Experts
The correct Answer is:
B
The magnetic field due to current in wire `1 ` in the region of wire `2` will be ` B_(1) = (mu_(0))/( 4 pi) ( 2i)/(b)` since wire `2` having current `i` is placed in a magnetic field ` B_(1)` , it will experience a force given by ` F = i(l B_(1) sin 90@)` :. force per unit length `(F)/(l) = i xx( mu_(0))/(4 pi) xx (2i)/(b) = ( mu_(0)i ^(2))/(2 pi b) ` [ :. B = (mu_(0))/(4 pi) xx ( 2 i)/(b)]`
Topper's Solved these Questions
MOVING CHARGES AND MAGNETISM
SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise MCQs(d )|1 Videos
SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise JEE Main And Advanced|209 Videos
Similar Questions
Explore conceptually related problems
Two thin long paralel wires separated by a distance b are carrying a current 1 A each. The magnitude of the force per unit length exerted by one wire on the other is
Two thin, long, parallel wires, separated by a distance 'd' carry a current of 'i' A in the same direction. They will
Two long straight parallel conductors separated by a distance of 0.5m carry currents of 5A and 8A in the same direction. The force per unit length experienced by each other is
Two long, parallel straight wires are separated by a distance of 0.2 cm and carry current of 10 A and 8 A. Find the magnitude of the attractive force exerted on 1 m length of one wire by the other wire.
Two long parallel wires, separated by a distance R have equal current I flowing in each of them. The magnetic field of one exerts a force F on the other. The distance R is increased to 2R and the current in each wire is reduced from I to I//2 . What is the force between them now?
SUNIL BATRA (41 YEARS IITJEE PHYSICS)-MOVING CHARGES AND MAGNETISM-MCQs(d )
