Question

what is the magnetic force value due to3.5 nano coloumb charge moves to theeast with a velocity of 250m/s makingangle of 30 degree with the magneticfiled(25m.T)?*

Answer 1

`2.2\cdot10^(-8)N`

Step-by-step explanation

The magnetic force due to the charge is given by:

`\begin{gathered} F=qv* B \\ \Rightarrow F=qvB\sin \theta \end{gathered}`

where q = electric charge

v = velocity of the charge

B = magnetic field

Therefore, substituting the given values into the equation, we have that the magnetic force due to the charge is:

`\begin{gathered} F=3.5\cdot10^(-9)\cdot250\cdot25\cdot10^(-3) \\ F=2.2\cdot10^(-8)N \end{gathered}`

That is the answer.