Question

A long straight wire carries a certain current and produces a
magnetic field of 2 × 10⁻⁴ weber/m² 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 [NEET Kar. 2013]
(charge on electron =1.6 × 10⁻¹⁹ C)
(a) Zero (b) 3.2 N
(c) 3.2 × 10⁻¹⁶ N (d) 1.6 × 10⁻¹⁶ N

Answer 1

Explanation:

Given magnetic field

B

^

=2×10

−4

Wbm

−2

velocity of electronv=10

7

m/s

charge of electron q=1.6×10

−19

C

We know magnetic forceF=qvBsinθ

F=1.6×10

−19

×10

7

×2×10

−4

=3.2×10

−16

Here θ=90 as mentioned electrons moving in a perpendicular direction to the wire viz. sin90=1

Answer 2

Answer:

its must be

Explanation:

b 3.2 N ......