Question

An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 10⁵m/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.

(1) 4 × 10^–20 N

(2) 8π× 10^–20 N

(3) 4π× 10^–20 N

(4) 8 × 10^–20 N

Answer 1

Magnetic field produced due to long current carrying wire at point A is

\(B =\frac{\mu_02i}{4\pi r}\)

\(= \frac{10^{-7}\times 2 \times 5}{20 \times 10^{-2}}\)

\(= \frac 12 \times 10^{-5}\)

Direction of magnetic field will be outward to the plane of paper.

Now, force acting on the electron due to this field,

\(\vec F = q(\vec v \times \vec B)\)

\(= 1.6 \times 10^{-19} \times 10^5 \times \frac 12 \times 10^{-5}\)

\(= 0.8 \times 10^{-19}\)

\(\therefore |\vec F| = 8 \times 10^{-20} N\)

Answer 2

Option : (4). 8 × 10^–20 N

f = 8 × 10^–20 Newton.