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Two long parallel wires carrying currents 2.5A and I amp in the same direction (directed into plane) are held at P and Q respectively as shown. The points P and Q are located at 5 m and 2m respectively from a collinear point R. An electron moving with a velocity of $4 \times 10^{5} m/s$ along positive x axis experiences a force of $3 .2 \times 10^{- 20} N$ at the point 'R'. The value of I is

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8 A

1 A

4 A

6 A

answer is A.

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Detailed Solution

F = qV (B_{1} + B_{2})

$3.2 \times 10^{- 20} = 1.6 \times 10^{- 19} \times 4 \times 10^{5} (\frac{μ_{0} 2.5}{2 π \times 5} + \frac{μ_{0} I}{2 π \times 2})$

$10^{- 20} = 2 \times 10^{- 10} \frac{μ_{0}}{2 π} (\frac{2.5}{5} + \frac{I}{2})$

$10^{- 6} = \frac{μ_{0}}{π} (0.5 + \frac{I}{2})$

$10^{- 6} = \frac{4 π \times 10^{- 7}}{π} (\frac{1}{2} + \frac{I}{2})$

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

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

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

$\Rightarrow I = 4 A$

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