Question

Two long parallel wires carrying currents 2.5 A and I A in the same direction (directed Into the plane of the paper) are held M P and Q respectively such that they are perpendicular to the plane of the paper. The points P and Q are located at a distance of 5 m and 2 m, respectively, from a collinear point R as shown in figure. An electron moving with a velocity of $4\times {10}^{5}m{s}^{-1}$ along the positive X-direction experiences a force of magnitude $3.2\times {10}^{-20}$ N at the point R.
The current I in wire Q is

• A. 1 A
• B. 2 A
• C. 3 A
• D. 4 A

Answer 1

Answer: (D)

4 A

$f=f_1+f_2$

$=qv{B}_1+qv{B}_2$

$=qv(\frac{\mu I_1}{2\pi R_1}+\frac{\mu I_2}{2\pi R_2})$

$\Rightarrow \frac{3.2}{{100.10}^{20}}=\frac{1.6\times 4\times {10}^{19}}{10}\times 2\frac{2}{{10}^7}(\frac{25}{5}+\frac{I}{2})$

$\frac{32}{16\times 4\times 2}\times 10=(\frac{1}{2}+\frac{I}{2})$

$\frac{5}{2}=\frac{1}{2}+\frac{I}{2}$

$2=\frac{I}{2}$

$\Rightarrow I=4A$

Hence, the answer is $4A.$