Question

A proton enters a magnetic field (B = 5.00 x 10^-2 T INTO THE PAGE) with

Answer 1

Answer:

r = 1.16 m

Explanation:

It is given that,

Magnetic field, $B=5.6\times 10^6\ m/s$

Velocity of proton, $v=5.6\times 10^6\ m/s$

We need to find the radius (in m) of the  circular path the particle will follow when it is travelling within the  magnetic field. The radius followed by the proton is given by :

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

m is mass of proton, $m=1.67\times 10^{-27}\ kg$

So,

$r=\dfrac{1.67\times 10^{-27}\times 5.6\times 10^6}{1.6\times 10^{-19}\times 5\times 10^{-2}}\\\\r=1.16\ m$

So, the radius of circular path is 1.16 m.