Question

A straight wire lying in a horizontal plane carries a current of 20 A. (Given, μ0 = 4π × 10-7 Tm/A)
(a) Find the magnitude of the magnetic field at a perpendicular distance of 35 cm from it on the plane. How do you know the direction of the magnetic field?
(b) If the length of the current carrying straight wire is 0.75m and it makes an angle of 60° with the Earth’s magnetic field 6× 10-5 T, find the force experienced by the wire.
(c) Now, ignore the earth’s magnetic field. The wire (carrying 20 A) feels an attractive force of 0.006 N/m due to another parallel wire which lies in the same plane carrying current 25 A. Calculate the distance between these two wires.

Answer 1

(a) Given the current through the wire is I = 20A and = 4 x 10-7Tm/A. The perpendcular distance from the wire to the point is r = 35cm = 0.35m. The magnetic field at a point at a distance r from an infinitely long straight wire is given by,

So the magnitude of the magnetic field is 1.14 x 10-5T.

The direction of this magnetic field can be found out using right hand thumb rule. According to this rule, hold the straight wire in the right hand such that the thumb indicate the direction of the current, then the curling fingers indicate the direction of magnetic field.

(b) If the length of the wire is known, it is is a finite wire. Given the length of the wire is = 0.75m, the earth's magnetic field is B = 6 x 10-5T and it makes an angle with the magnetuc filed.The force on the wire is given by,

So the force on the wire is 7.80 x 10-4N.

(c) Let the current through the first wire be I1 = 20A and that though the second wire be I2 = 25A. Let r be the distance between the wires. The attractive force per unit length between these wires is F = 0.006N/m. The attractive force between these wires is given as,

So the distance between these wires is 1.67cm.