Two infinitely long straight wires A and B, each carrying current I are placed on x and y-axis, respectively. The current in wires A and B flow along `-hat(i)` and `hat(j)` directions respectively. The force on a charge particle having vharge q, moving from position, `r=d (hat(i)+hat(j))` with velocity `v= vhat(i)` is

Options

A. `(mu_(0) lq v)/(2 pi d) hat(j)`
B. `(mu_(0)lq v)/(pi d) hat(j)`
C. `(mu_(0) l q v)/(sqrt(2) pi d) hat(k)`
D. 0

Answer

B

Explanation

The given situation for two curent carying wires are shown in the figures.

Position vector of charge particle,
`r=d(hat(i)+hat(j))=d hat(i)+d hat(j)`
The magnetic field intensity due to the wire along `-hat(i)` direction,
`B_(1)=mu_(0)/(2pi). I/d (-hat(k))`
Similarly, magnetic field intensity due to wire along `hat(j)` direction.
`B_(2)=mu_(0)/(2pi).I/d (-hat(k))`
`:.` Resultant magnetic field at point P due to both the wires,
`B=B_(1)+B_(2)=2.mu_(0)/(2pi).I/d(-hat(k))`
`B=mu_(0)/pi .I/d (-hat(k))`
`:.` Force on the charged particles.
`F=q(v xx B)=q[v hat(i) xx mu_(0)/pi I/d. (-hat(k))]=(mu_(0)I qv)/(pi d). hat(j)`