Question

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.

Answer 1

Answer:

1. The magnitude of magnetic field is directly proportional to the magnitude of current through the loop. i.e. B∝I.

2. The magnitude of magnetic field is inversely proportional to the radius of the circular loop. i.e. B∝1r.

Explanation:

hence,  given      

                    r=20cm,  I=10A ,    V=2.0 x 10^6 m s-1

             therefore,

         B=(permeability of free space) x I  /  2пr

           B= (4п x 10^-7  T * M/A ) X 10 A /  2*п*20cm

          B=10^-7 T (tesla)

           F=qvB sinθ,

           F=1.6*10^-19 * 2.0 x10^6 x 10^-7 x  (sin30)degree

             F=1.6 * 10^-20  N

   

          I hope it helps..

Answer 2

The magnitude of the magnetic force on the electron at the instant it crosses the plane is $16 \pi \times 10^{-19} N$

Explanation:

Given data in the question  

In the loop, current magnitude, I = 10 A

Loop Radius r = 20 cm = $20 \times 10^{-2} m$

The magnetic field strength at the core is therefore determined

$B=\frac{\mu_{0} i}{2 r}$

Electron velocity  v = 2 × $10^{6}$ m/s

Angle between velocity and intensity of magnetic field, θ = 30°

The magnetic force on the electron is thus given by the

$F=e v B \sin \theta$

   $=1.6 \times 10^{-19} \times 2 \times 10^{6} \times \frac{\mu_{0} i}{2 R} \sin 30^{\circ}$

   $=1.6 \times 10^{-19} \times 2 \times 10^{6} \times \frac{\left(4 \pi \times 10^{-7} \times 10\right)}{2 \times 20 \times 10^{-2}} \times \frac{1}{2}$

   $=3.2 \times 10^{-19} \times 10^{6} \times \frac{\left(4 \pi \times 10^{-7} \times 10\right)}{2 \times 20 \times 10^{-2}} \times 0.5$

   $=3.2 \times 10^{-13} \times \frac{\left(40 \pi \times 10^{-7}\right)}{40 \times 10^{-2}} \times 0.5$

   $=3.2 \times 10^{-13} \times \frac{\left(\pi \times 10^{-7}\right)}{10^{-2}} \times 0.5$

   $=3.2 \times 10^{-13} \times \frac{\left(\pi \times 10^{-7}\right)}{10^{-2}} \times 0.5$

   $=1.6 \times 10^{-13} \times \frac{\left(\pi \times 10^{-7}\right)}{10^{-2}}$

   $=16 \pi \times 10^{-19} N$

The magnitude of the magnetic force on the electron is $16 \pi \times 10^{-19} N$