Three distinct current carrying wires intersect a finite rectangular plane ABCD. The current in left wire and the loop is `I_1`. The direction of current in left most wire and right most loop is downwards as shown in figure. The current `I_2` through middle wire is adjusted so that the path integral of the total magnetic field along the perimeter of the rectangle is zero, that is, `oint_(ABCDA) vec B. vec (dl) = 0`. Then the current `I_2` is-


A. `2I_(1)` and upwards
B. `2I_(1)` and downwards
C. `4I_(1)` and upwards
D. `3I_(1)` and downwards

Options

A. `2I_(1)` and upwards
B. `2I_(1)` and downwards
C. `4I_(1)` and upwards
D. `3I_(1)` and downwards

Answer

C

Explanation

For calculated of `oint_(ABCD)vec(B).vec(dl)` current upwards shall be positive and current downwards shall be negative Therefore, `oint vec(B).vec(dl)=(I_(2)-I_(1))+I_(2)-4I_(1)=0`
Therefore `I_(2)=4I_(1)` upwards