Answer to Question #223037 in Electricity and Magnetism for Stan

Question

A proton moves at 8x10^6 m/s along the x axis. It enters a region in which there is a magnetic field of magnitude 2.5 T directed at an angle of 60.0 degrees with the y axis and laying on the x-y plane. Find A) The magnitude and direction of the force on the proton B) The protons initial acceleration C) Is there any change in magnitude and direction and the initial acceleration if the charged particles are electron instead of proton

Answer 1

$F=q(V\times B)=qVBsin\theta$

$$F=1.6\times10^{-19}\times8\times10^8\times2.5\times sin60°=3.2\times10^{-10}\times\frac{\sqrt{3}}{2}N$$

F=27.72N

Direction right hand palm law

F direction is upward direction

B. When charge particle is proton

$ma=qvBsin60°$

$a=\frac{qvB sin60°}{m}$

$$a=\frac{1.6\times10^{-19}\times2.5\times\frac{\sqrt{3}}{2}}{1.67\times10^{-27}}=2.08\times10^8m/sec$$

C. When charge particle is electron

$$q=1.6\times10^{-19}C,m_e=9.1\times10^{-31}kg$$

$ma=qvBsin60°$

$a=\frac{qvB sin60°}{m}$

$$a=-\frac{1.6\times10^{-19}\times2.5\times\frac{\sqrt{3}}{2}}{9.1\times10^{-31}}=-3.80\times10^{11}m/sec^2$$