Question

Question 4.20: A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10 5 V m −1 , make a simple guess as to what the beam contains. Why is the answer not unique?

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-171

Answer 1

Narrow beam of charged particles remains undeflected and is perpendicular to both electric and magnetic field which are mutually perpendicular. so, the electric force is balanced by magnetic force.
e.g., $qE=Bqv$
so, speed of charged particles , v = E/B
here, E = 9 × 10^-5 V/m and B = 0.75T
so, v = 9 × 10^-5/0.75 = 12 × 10^-5 m/s

because the beam is accelerated thri515kV, if charge is q , then kinetic energy gained by charged particles.
$\frac{1}{2}mv^2 =qV\\\\\frac{m}{q}=\frac{2V}{v^2}$
= (2 × 15 × 10³)/(12 × 10^-5)²
= 20.8 × 10¹¹ kg/C
Here, we can only obtain charge to mass ratio and same ratio can be in Deuterium ions , He++, Like+++ so, the beam can obtain any of these charged particles .

Answer 2

Hey !!

Given

        B = 0.75 T

       E = 9 × 10⁵ Vm⁻¹

      V = 15 kV = 15000 V

The velocity of electron v is given by

                $\frac{1}{2} mv^{2} = eV$ (or)  v = √2eV/m

Substituting value of V, we get

         v = √2e × 15000 / m      = √3 × 10⁴(e/m)

If particles are undeflected in simultaneous transverse electric and magnetic field,   eB = evB

              v =E/B    =>    √3 × 10⁴ e/m = E/B

(OR)             e/m = E²/B² × 1/3 × 10⁴⁴

   = (9 × 10⁵)² / (0.75)² × 1/3 × 10⁴

=> FINAL RESULT = 4.8 × 10⁷ C/kg

GOOD LUCK !!