Answer to Question #174099 in Electricity and Magnetism for ray

Question

An ion with a +1 charge is fired through a 150 mT magnetic field as shown below. The particle is moving to the right at a speed of 7.1e+6 m/s. What is the magnitude and direction of the force on the charge?

Answer 1

Given,

Charge on the ion $(q)= 1C$

magnetic field $(B)=$ $150\times 10^{-3}T$

Velocity of the ion $v=7.1\times 10^{6}m/s$

So, force on the ion $(F)=qv\times B$

$=7.1\times 10^{6}\times150\times10^{-3}N$

$=1065\times 10^{3}N$

$=1.065\times 10^6 N$

As per the right hand thumb rule,

Magnetic force will be perpendicular to the direction of motion of charged ion and direction of magnetic field both.

As per the question's instruction, diagram must be attached but here diagram is not visible, so i am taking the assumption.