Question

A magnetic field of 3.6 T [down] is 3.0 m wide. Find the magnetic force on a proton moving at 4.2 × 10^5 m/s [east] in the field. Charge of a proton = 1.6x10-19C

Answer 1

$2.41*10^{-13}\text{ Newtons}$

Explanation

to solve this we will use the formula

$\begin{gathered} v=\frac{F}{qB} \\ where\text{ v is the velocity} \\ F\text{ is the force} \\ q\text{ is the charge} \\ B\text{ is the magnetic field} \end{gathered}$

so

Step 1

a)let

$\begin{gathered} B=3.6\text{ T} \\ q=1.6*10^{-19}C \\ v=4.2*10^5\text{ m/s} \end{gathered}$

b) now, replace and solve for F

$\begin{gathered} v=\frac{F}{qB} \\ Multiply\text{ both sides by qB} \\ F=v*q*B \\ F=4.2*10^5\frac{m}{s}*1.6*10^{-19}C*3.6 \\ F=2.41*10^{-13}\text{ N} \end{gathered}$

therefore, the answer is

$2.41*10^{-13}\text{ Newtons}$

I hope this helps you