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
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:
[tex]r=\frac{mv}{qB}[/tex]
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:
[tex]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}[/tex]
Then, by replacing in the formula for the radius you obtain:
[tex]r=\frac{(1.67*10^{-27}kg)(6.92*10^5\frac{m}{s})}{(1.6*10^{-19}C)(0.60T)}=0.012m[/tex]
hence, the radius is 0.012m
