Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# A Circular Loop of Radius 20 Cm Carries a Current of 10 A. an Electron Crosses the Plane of the Loop with a Speed of 2.0 × 106 M S−1. - Physics

Short Note

A circular loop of radius 20 cm carries a current of 10 A. An electron crosses the plane of the loop with a speed of 2.0 × 106 m s−1. The direction of motion makes an angle of 30° with the axis of the circle and passes through its centre. Find the magnitude of the magnetic force on the electron at the instant it crosses the plane.

#### Solution

Given:
Magnitude of current in the loop, I = 10 A
Radius of the loop, r = 20 cm = 20 × 10−2 m
Thus, the magnetic field intensity at the centre is given by $B = \frac{\mu_0 I}{2r}$

Now,
Velocity of the electron, v = 2 × 106 m/s
Angle between the velocity and the magnetic field intensity, θ = 30°
Thus, the magnetic force on the electron is given by

$F = evB\sin \theta$

$= 1 . 6 \times {10}^{- 19} \times 2 \times {10}^6 \times \frac{\mu_0 i}{2R}\sin 30^\circ$

$= 1 . 6 \times {10}^{- 19} \times 2 \times {10}^6 \times \frac{4\pi \times {10}^{- 7} \times 10}{2 \times 20 \times {10}^{- 2}} \times \frac{1}{2}$

$= 16\pi \times {10}^{- 19}$ N

Concept: Magnetic Field on the Axis of a Circular Current Loop
Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 13 Magnetic Field due to a Current
Q 38 | Page 252