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}`

Step-by-step explanation

to solve this we will use the formula

`\begin{gathered} v=(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=(F)/(qB) \\ Multiply\text{ both sides by qB} \\ F=v*q*B \\ F=4.2*10^5(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