Question

PHYS 2523 (UNIVERSİTY Puvsos msu (20 pts/S pt. each) As shown in the fiqgure, the magnetic field over a certain range is eiven by B B,k, where B,- 0.ST. A proton moves into the magnetie field with a velocity i++v,k, where v, 1 x 10 m/s, Vy 3x 10% m/s, and v 3 x 10* m/s. The charge of the proton is 1.60 x 10" C, and the mass of the prokon is 1.67 x 10 kg. Here t.J.k are unit vectors in the x, y and z directions, loy electon when placed w harged pariele moves in a wre Force on a Electric fields: (a) Find the magnetic force on the proton (F). Magpeie felds. Lorenta toree: Force on a curvenk s Magvetie fleld A long suraig Ai the cente Along the Al the ce (b) What is the magnitude of the force acting on it (Fa? .Am Bw (c) Find the magnitude of the acceleration of the proton (d) Find i, and h(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

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