Two parallel, long wires carry current `i_(1)` and `i_(2)` with`i_(1)gti_(2)`. When the currents are in the same direction, the magnetic field at a point midway between the wires is `30 muT`. If the direction of `i_(1)` is reversed , the field becomes `90muT`. The ratio `i_(1)//i_(2)` is

Options

A. 4
B. 3
C. 2
D. 1

Answer

c

Explanation

`(c) B=(mu_(0)i)/(2pid),B propi`.
`B_(0)=-(mu_(0)i_(1))/(2pid)+(mu_(0)i_(2))/(2pid)=90xx10^(-6)......(1)`
If direction is reversed in wire `(1)`
`B_(0)^(')=+(mu_(0)i_(1))/(2pid)+(mu_(0)i_(2))/(2pid)30xx10^(-6) ....(2)`
From `eq (1)` and `(2) (mu_(0)I_(2))/(pid)=120xx10^(-4),(mu_(0)I_(1))/(pid)=60xx10^(-6)implies(I_(2))/(I_(1))=2`