Two long and parallel straight wires A and B , carrying currents of 8 . 0 A and 5 . 0 A in the same direction, are separated by a distance of 4 . 0 cm . Estimate the force on 10 cm section of wire A .

Here, the two wires A and B are long and parallel straight wires. The gap between them is of d = 4 cm = 0 . 04 m Current passing through wire A is I A = 8 A and through wire B is I B = 5 A . We have to calculate the force on 10 cm section of wire A. Thus, let L = 10 cm = 0 . 1 m . So the force on 10 cm section of wire A : F B A = μ ∘ 2 I A I B L 4 π d F B A = 4 π × 10 - 7 × 2 × 8 × 5 × 0 . 1 4 π × 0 . 04 N F B A = 2 × 10 - 5 N . Hence, 2 × 10 - 5 N of force is acting on the 10 cm section of wire A.

State and prove Ampere's circuital law.

Ampere's circuital law state that line integral of magnetic field forming a close loop around the current (i) carrying wire in the plane normal to the current, is equal to the μ o times the net current passing through the close loop. ϕ B . d l = μ o i B = μ o i 2 πr , 1 / d l = 2 πr B / dl = μ o i 2 πr × 2 πr = μ o i so B / dl = μ o i

A galvanometer coil has a resistance of 15 Ω . The meter shows full scale deflection for a current of 4 mA . How will you convert the galvanometer into ammeter of range 0 to 6 A ?

Here R g = 15 Ω , I g = 4 m A = 0 .004 A , I = 6 A R s = I g I − I g × R g = 0 .004 6 − 0 .004 × 15 = 0 .010 Ω = 10 mΩ By connecting a shunt of resistance 10 mΩ across the given galvanometer, we get an ammeter of range 0 to 6 A .

What is Biot-Savart's law?

Biot-Savart's law states that the magnetic field produced at a point near a current-carrying conductor is proportional to the material of the medium μ 0 , the current II flowing in the conductor, the small length of the wire dl involved, and inversely proportional to distance r between the point and the conductor.