Two long and parallel straight wires A and B carrying currents of8.0 A and 5.0 A in the same direction are separated by a distance of4.0 cm. Estimate the force on a 10 cm section of wire A.
Given: The current carried by the wire A is 8 A , the current carried by the wire B is 5 A and the distance between the wires is 10 cm .
Force exerted on wire A due to magnetic field is given as,
F = μ 0 2 I A I B l 4 π r
where, the permeability of free space is μ 0 , current carried by wire A is I A , current carried by wire B is I B , distance between A and B is r .and the section of wire A is l
By substituting the given values in the above equation, we get,
F = 4 π × 10 − 7 × 2 × 8 × 5 × 0.1 4 π × 0.04 = 2 × 10 − 5 N
Therefore, the force exerted on wire A due to magnetic field is 2 × 10 − 5 N .
Given: The current carried by the wire
Force exerted on wire
where, the permeability of free space is
By substituting the given values in the above equation, we get,
Therefore, the force exerted on wire
