An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of

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Solution

Q.
An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of $10^{5}$ $m/s$ parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

Electron $v = 10^{5}$ $m/s$ [2021]

1 $8 \times 10^{- 20}$ $N$

2 $4 \times 10^{- 20}$ $N$

3 $8 π \times 10^{- 20}$ $N$

4 $4 π \times 10^{- 20}$ $N$

Ans.
(1)

The magnetic field due to a straight wire at the location of the electron is

$B = \frac{μ_{0}}{4 π} \cdot \frac{2 I}{r}$

$B = 10^{- 7} \times \frac{2 \times 5}{0.2} = 50 \times 10^{- 7} T (\hat{k})$

The direction is given by right-hand thumb rule.

The force on a charged particle moving in a magnetic field is

$\vec{F} = q (\vec{v} \times \vec{B})$

$\vec{F} = - 1.6 \times 10^{- 19} \times (10^{5} \hat{i} \times 50 \times 10^{- 7} \hat{k})$

$\vec{F} = 8 \times 10^{- 20} N \hat{j}$

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