Question

one proton beam enter magnetic feild of 10^-4T normally ,specific charge=10^11C/Kg.velocity=10^7m/s.what is te radius of circle

Answer 1

Explanation:

The radius of the proton in the circular path is given by:

r=

qB

mv

=

10

11

×10

−4

10

7

=1m(∵q/m=10

11

C/kg)