Question

A 3.4 nC charged particle has a velocity of 4.7 m/s and is moving in the +x-direction. If this charge is in a magnetic field determined by B =-1.4 T t + 7.52T), what is the magnetic force on this particle?

Answer 1

Answer:

Magnetic force, $F=1.22\times 10^{-7}\ N$

Explanation:

It is given that,

Charge, $q=3.4\ nC=3.4\times 10^{-9}\ C$

Velocity, v = 4.7 m/s

Magnetic field, $B=-1.4i+7.52j$

$|B|=\sqrt{(-1.4)^2+(7.52)^2}=7.64\ T$

Magnetic force is given by :

$F=q\times v\times B$

$F=3.4\times 10^{-9}\ C\times 4.7\ m/s\times 7.64\ T$

$F=1.22\times 10^{-7}\ N$

So, the magnetic force on this particle is $1.22\times 10^{-7}\ N$. Hence, this is the required solution.