Question

Two long parallel wires carrying currents 2.5 A and "I" A in the same direction (directed in the o plane) are held at P and Q respectively as shown. The points P and Q are located at a distance of 5 m and 2 m respectively from a collinear point R. An electron moving with a velocity of $4\times {10}^{5}m{s}^{-1}$ along the positive x-axis experiences a force of magnitude $3.2\times {10}^{-20}N$ at the point R. The value of I is

• A. 8 A
• B. 6 A
• C. 4 A
• D. 2 A

Answer 1

The correct option is C

4 A

Resultant force on the electron,
$F=qvB=qv(B_1+B_2)\Rightarrow F=qv(\frac{{\mu }_0\times 2.5}{2\pi \times 5}+\frac{{\mu }_0\times I}{2\pi \times 2})\Rightarrow 3.2\times {10}^{-20}=1.6\times {10}^{-19}\times 4\times {10}^5\times (\frac{4\pi \times {10}^{-7}\times 2.5}{2\pi \times 5}+\frac{4\pi \times {10}^{-7}\times I}{2\pi \times 2})$
Solving we get, $I=4A$