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

Magnetic force,
`F=1.22* 10^(-7)\ N`

Step-by-step explanation:

It is given that,

Charge,
`q=3.4\ nC=3.4* 10^(-9)\ C`

Velocity, v = 4.7 m/s

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

`|B|=√((-1.4)^2+(7.52)^2)=7.64\ T`

Magnetic force is given by :

`F=q* v* B`

`F=3.4* 10^(-9)\ C* 4.7\ m/s* 7.64\ T`

`F=1.22* 10^(-7)\ N`

So, the magnetic force on this particle is
`1.22* 10^(-7)\ N`. Hence, this is the required solution.