Chapter 4: Q2E (page 912)

A particle of mass 0.195 g carries a charge of $- 2.50 \times 10^{- 8} C$ . The particle is given an initial horizontal velocity that is due north and has magnitude $4.00 \times 10^{4} m / s$ . What are the magnitude and direction of the minimum magnetic field that will keepthe particle moving in the earth’s gravitational field in the samehorizontal, northward direction?

Short Answer

Expert verified

The Magnetic field is $B_{E} = 1.91 T$

Step by step solution

Important Concepts

The magnetic field force is given by

$F_{B} = q (v \times B) F = q v B \sin θ$

Where q is the charge of the particle, V is the velocity and B is the magnetic field

Application

In order to keep the particle moving in the same direction, the particle should be in equilibrium I.e the magnetic field force to be equal to the gravitational force in magnitude but opposite in direction.

According to the right hand rule and that the particle contains negative charge, the direction of the magnetic field must be towards east.

Hence,

$m \overset{⇀}{g} = q (\overset{⇀}{v} \times \overset{⇀}{B}) m \overset{⇀}{g} = q |V| |B| B = \frac{m \overset{⇀}{g}}{q |V|}$

Substitute all value in the above equation.

$B = \frac{0.195 \times 10^{- 3} k g \times 9.81 m / s^{2}}{2.5 \times 10^{- 8} C \times 4 \times 10^{4} m / s} B = 1.91 T Hence the magnitude of the field is B_{E} = 1.91 T and the direction is east$