Question

A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60-T magnetic field. What is the radius of the resulting path

Answer 1

Answer:

0.012m

Explanation:

To find the radius of the path you can use the formula for the radius od the trajectory of a charge that moves in a constant magnetic field:

$r=\frac{mv}{qB}$

m: mass of the proton 9.1*10^{-31}kg

q: charge of the proton 1.6*10^{-19}C

B: magnitude of the magnetic field 0.60T

v: velocity of the proton

In order to use the formula you need to calculate the velocity of the proton. This can be made by using the potential difference and charge, that equals the kinetic energy of the proton:

$qV=E_k=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.5*10^{3}V)}{1.67*10^{-27}kg}}=6.92*10^{5}\frac{m}{s}$

Then, by replacing in the formula for the radius you obtain:

$r=\frac{(1.67*10^{-27}kg)(6.92*10^5\frac{m}{s})}{(1.6*10^{-19}C)(0.60T)}=0.012m$

hence, the radius is 0.012m

Answer 2

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm