Three long, D straight parallel wires, carrying current, are arranged as shown in figure. The force 30 A 10 A 20A experienced by a 25 cm length of wire C is

Options

A. `10^(-3) N`
B. `2.5 xx 10^(-3)N`
C. zero
D. `1.5x X 10^(-3) N`

Answer

c

Explanation

The magnetic field D C G due to wire D at wire C
`B_(D)=(mu_(0)/(4pi))(2I)/(r)`
`=(10^(-7)xx2xx30)/(0.03)=2xx10^(-4)T`
which is directed into the page.
The magnetic field due to wire G at C is
`B_(G)=(10^(-2)xx2xx20)/(0.02)=2xx10^(-4)T`
which is directed out of the page. Therefore, the field at the position of the wire C
`B=B_(D)-B_(G) = 2 xx 10^(-4) - 2 xx 10^(-4) = zero`
The force on 25 cm of wire C is `F=Bllsin theta =zero`