00:01
Here for the solution for the step one.
00:04
The expression for magnetic force that is f vector on a charge particle of charge q moving with a velocity v vector in a magnetic field b vector is given as following equation that f vector equal to q in bracket v vector multiply by b vector.
00:21
Using above expression for magnetic force magnitude of force f is given by the following expression that f equal to q v b -sign theta.
00:32
Here, theta is the angle between velocity of the charge particle and magnetic field.
00:38
Now, the magnetic force f on an electron when it is moving in a magnetic field b, with speed v, has maximum values when the electrons moving perpendicular to the field.
00:53
Thus, the maximum force on an electron is f max equal to qvb.
00:58
Now rearrange the above equation for b we get b equal to f max divided by qv now substitute 8 .2 multiply by 10 to the power minus 13 newton for f max 1 .6 multiply by 10 to the power minus 19 for c 1 .6 multiply by 10 to the power minus 19 column for q and 2 .8 multiply by 10 to the power 6 meter per second for v in the above equation by substituting value we get b equal to 8 .2 multiply by 10 to the power minus 13 newton divided by 1 .6 multiply by 1020 .19 column in bracket 2 .8 multiplied by 10 to 2 .8 minus 6 meter per second.
01:37
And here by solving this, we get the value of b...