Two parallel wires in the plane of the paper are distance `X_(0)` apart. A point charge is moving with speed `u` between the wires in the same plane at a distance `X_(1)` from one of the wires. When the wires carry current of magnitude `I` in the same direction , the radfius of curvature of the path of the point charge is `R_(1)`. In contrast, if the currentsd `I` in the two wires have directions opposite to each other, the radius of curvature of the path is `R_(2)`. if `(X_(0))/(X_(1)) = 3`, the value of `( R _(1))/( R_(2))` is

Options

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

Answer

C

Explanation

when currents in the two wires are in the same direction, then magnetic field induction at O is

`B_1=(mu_0I)/(2pix_1)-(mu_0I)/(2pi(x_0-x_1))`
`=(mu_0I)/(2pi)[1/x_1-(1)/((x_0-x_1))]=(mu_0I)/(2pi)[(x_0-2x_1)/(x_1(x_0-x_1))]`
When the currents in the two wires are in opposite directions, then
`B_2=(mu_0I)/(2pi)[1/x_1+(1)/((x_0-x_1))]=(mu_0I)/(2pi)[(x_0)/(x_1(x_0-x_1))]`
In magnetic field, the radius of curvature of curved path of the moving charged particle is related to B as
`Rprop1/B`
or `R_1/R_2=B_2/B_1=(x_0)/((x_0-2x_1))=(x_0//x_1)/((x_0//x_1-2))=(3)/(3-2)=3`