Question

Plz answer all questions

(20 pts/5 pt. each) As shown in the figure, the magnetic field over a certain range is given by B- B,k, where B,-0.5T.A proton moves into the magnetic field with a velocity - i+v,j+v,k, where v,-1x 10 m/s, By 3 x 10s m/s, and v, 3x10s m/s. The charge ofthe proton is 160x10" C,and the mass ofthe proton is 1.67 x 10 kg Here t.J.k are unit vectors in the x, y and z directions, respectively 5. (a) Find the magnetic force on the proton (F (b) What is the magnitude of the force acting on it (Fp? (c) Find the magnitude of the acceleration of the proton (d) Find , and y (You can use the figure). Then, find the radius of the resulting helix

Answer 1

The magnetic field is given by

$\overrightarrow{B}=0.5T\hat{k}$

The velocity of the proton is given by

$\overrightarrow{v}=(\hat{i}+3\hat{j}+3\hat{k})\times 10^{5}m/s$

the mass of the proton is

$m=1.67\times 10^{-27}Kg$

The charge of the proton is

$q=1.6\times 10^{-19}C$

The magnetic force is given by

$\overrightarrow{F}_{B}=q(\overrightarrow{v} \times \overrightarrow{B})$

or,

$\overrightarrow{F}_{B}=1.6 \times 10^{-19}( ((\hat{i}+3\hat{j}+3\hat{k})\times 10^{5} )\times 0.5\hat{k})$

or,

$\overrightarrow{F}_{B}=2.4\times 10^{-14}N\hat{i}-0.8\times 10^{-14}N\hat{j}$

b)

the magnitude of the magnetic force is given by

$F_{B}=\sqrt{(2.4\times 10^{-14})^2+(0.8\times 10^{-14})^2}=2.53\times 10^{-14}N$

c)

The magnitude of the acceleration of the proton is given by

$a=\frac{F_{B}}{m}=\frac{2.53\times 10^{-14}}{1.67\times 10^{-27}}=1.52\times 10^{13}m/s^{2}$

d)

$v_{\perp }=\sqrt{(10^{5})^2+(3\times 10^{5})^2}=3.16\times 10^{5}m/s$

$v_{\parallel }=3\times 10^{5}m/s$

The radius of the resulting helix is given by

$l=2\pi \frac{mv_{\parallel }}{qB}=2\pi =2\pi \times \frac{1.67\times 10^{-27}\times 3\times 10^{5}}{1.6 \times 10^{-19}\times 0.5}=0.04m=4cm$