A long straight wire carries a certain current and produces a magnetic field 2 × 10-4 Wb m 2 at a perpendicular distance of 5 cm from the wire. An electron situated at 5 cm from the wire moves with a velocity 107 m / s towards the wire along perpendicular to it. The force experienced by the electron will be (charge on electron 1.6 × 10-19 C ) :

Question

A long straight wire carries a certain current and produces a magnetic field 2 × 10-4 Wb m 2 at a perpendicular distance of 5 cm from the wire. An electron situated at 5 cm from the wire moves with a velocity 107 m / s towards the wire along perpendicular to it. The force experienced by the electron will be (charge on electron 1.6 × 10-19 C ) : (A) 3.2 NR (B) 3.2×10-16N (C) 1.6×10-16N (D) zero

Tardigrade Admin · Accepted Answer

The situation is as shown in the figure.

Here,

v = 1 0^{7} m / s, B = 2 \times 1 0^{- 4} Wb / m^{2}

The magnitude of the force experienced by the
electron is

F = e v B sin θ

(∴ v

and

B

are perpendicular to each other)

= e v B sin 9 0^{\circ}

= 1.6 \times 1 0^{- 19} \times 1 0^{7} \times 2 \times 1 0^{- 4} \times 1

= 3.2 \times 1 0^{- 16} N

$3.2 NR$

$3.2 \times 1 0^{- 16} N$

$1.6 \times 1 0^{- 16} N$

zero

3.2NR

3.2×10−16N

1.6×10−16N

zero

The situation is as shown in the figure. Here, v=107m/s,B=2×10−4Wb/m2 The magnitude of the force experienced by the electron is F=evBsinθ (∴v and B are perpendicular to each other) =evBsin90∘ =1.6×10−19×107×2×10−4×1 =3.2×10−16N

$3.2 NR$

$3.2 \times 1 0^{- 16} N$

$1.6 \times 1 0^{- 16} N$