An infinitely long straight conductor carries a current of 5 A . An electron is moving with a speed of 10^5 m/

Question

An infinitely long straight conductor carries a current of

5 A

. An electron is moving with a speed of

1 0^{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 the instant.

Accepted Answer

Magnetic field produced by the conductor at distance

r = 20 cm = 0.2 m

:

B = \frac{μ _{0} I}{2 π r} = \frac{4 π \times 1 0 ^{- 7} \times 5}{2 π \times 0.2} = \frac{2 \times 1 0 ^{- 7} \times 5}{0.2} = 5 \times 1 0^{- 6} T

The magnetic force on the electron (

q = 1.6 \times 1 0^{- 19} C

):

F = q v B sin θ

Since the velocity is parallel to the conductor and the field is circular (perpendicular to velocity),

θ = 9 0^{\circ}

.

F = (1.6 \times 1 0^{- 19}) \times (1 0^{5}) \times (5 \times 1 0^{- 6})

F = 8.0 \times 1 0^{- 20} N

Solution & Step-by-step Explanation