Question

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 charged particle having charge $q$, moving from position, $r =d(\hat{ i }+\hat{ j })$ with velocity $v =v \hat{ i }$ is

• A. $\frac{\mu_{0} l q v}{2 \pi d} \hat{ j }$
• B. $\frac{\mu_{0} l q v}{\pi d} \hat{ j }$
• C. $\frac{\mu_{0} lq v}{\sqrt{2} \pi d} \hat{ k }$
• D. 0

Answer 1

Answer: (B)

The given situation for two current carrying wires are shown in the figures.

Position vector of charge particle,
$r =d(\hat{ i }+\hat{ j })=d \hat{ i }+\hat{d j }$
The magnetic field intensity due to the wire along $-\hat{ i }$ direction,
$B _{1}=\frac{\mu_{0}}{2 \pi} \cdot \frac{I}{d}(-\hat{ k })$
Similarly, magnetic field intensity due to wire along $\hat{ j }$ direction.
$B _{2}=\frac{\mu_{0}}{2 \pi} \cdot \frac{I}{d}(-\hat{ k })$
$\therefore $ Resultant magnetic field at point $P$ due to both the wires,
$B = B _{1}+ B _{2}=2 \cdot \frac{\mu_{0}}{2 \pi} \cdot \frac{I}{d}(-\hat{ k }) $
$B =\frac{\mu_{0}}{\pi} \cdot \frac{I}{d}(-\hat{ k })$
$\therefore $ Force on the charged particles,
$F =q( v \times B )=q\left[v \hat{ i } \times \frac{\mu_{0}}{\pi} \frac{I}{d} \cdot(-\hat{ k })\right]=\frac{\mu_{0} I \,q v}{\pi d} \cdot \hat{ j }$