Question

A proton enters a uniform magnetic field of 5t with velocity 4 into 10 to the power 7 m per second at right angle to the field the magnetic force acting on the proton is

Answer 1

we have formula F=qvBsintheta

given magnetic field = 5 Tesla

v=4*10^7 m/s

q=1.6*10^-19 C

F=1.6*10^-19*4*10^7*5*sin 90 (angle between v and B is 90 given in question itself)

F=1.6*5*4*10^-12

F=32*10^-12 N

hope this helps

Answer 2

Given:

A Proton enters a magnetic field of $5T$ with a velocity $4*10^7m/s$ at a right angle to the field.

To FInd:

The magnetic force acting on the proton, $F=?$

Solution:

The magnetic force acting on a charge $q$ moving with a velocity $v$ inside a magnetic field $B$

                                     $F=qBvsin\alpha$

directed perpendicular to the plane containing $B$ and $v$, where $\alpha$ is the angle between  $B$ and $v$.

Charge of a proton, $q=1.6*10^{-19}C$

                         ∴$F=1.6*10^{-19}*5*4*10^7sin90^o$

                       ⇒ $F=3.2*10^{-11}N$

Hence, the magnitude of the force acting on the proton inside the magnetic field, $F=3.2*10^{-11}N$