00:01
In this problem, we are given that there is an electron beam which is moving with a speed of 3 times 10 raised to 5 meter per second in a direction perpendicular to the magnetic field and the region is containing uniform magnetic field where the field strength is 0 .75 tesla and we are required to determine the magnitude of the force that's experienced by this charged particle which is moving.
00:30
In the region of uniform magnetic field, and along with that we have to also compute the radius of the trajectory that's followed by this charged particle.
00:40
So here, the electron is moving perpendicular to the magnetic field, so the angle theta between velocity vector and magnetic field vector is 90 degrees.
00:51
And to compute the magnitude of the force, let's use this equation, according to which if we take the magnitude it's bqv sine theta so let's substitute the values here b is 0 .75 times 3 into 10 raise to 5 multiplied with the charge carried by the particle which is a proton that's given so it's 1 .6 times 10 raise to minus 19 and this is multiplied with sine 90 degrees.
01:24
So when we multiply here 0 .75 with 3 times 10 raised to 5 and that with 1 .6 times 10 raised to minus 19 and sine 90 that's anyways going to be 1.
01:37
So we get the magnitude of the force coming out to be 3 .6 times 10 raise to minus 14 newtons and to compute the radius of curvature we use the idea that the magnetic force that's applied on the electron it gives rise to the centripetal force which is given by this equation which is mv square.
01:58
So when we equate both of them, we get bqv equal to mv square by r...