A proton, which moves perpendicular to a magnetic field of 1.35 T in a circular path of radius 0.072 m, has what speed? (qp = 1.6 · 10 19 C and mp = 1.67 · 10 27 kg)

Question

asked Sep 11, 2020 128k views

1 Answer

Niko Jojo · Answer 1 · 2020-09-16T05:23:06+0000

Answer:

Speed of the proton is
$v=9.31* 10^6\ m/s$

Step-by-step explanation:

It is given that,

Magnetic field, B = 1.35 T

Radius of circular path, r = 0.072 m

The centripetal force is balanced by the magnetic force such that,

(mv^2)/(r)=qv

v=(qrB)/(m)

q and m is the charge and mass of a proton respectively

v=(1.6* 10^(-19)* 0.072* 1.35)/(1.67* 10^(-27))

$v=9.31* 10^6\ m/s$

So, the speed if the proton is
$9.31* 10^6\ m/s$ . Hence, this is the required solution.