00:01
In this problem, we have been given that there is a velocity selector and we need to determine the magnetic field of this velocity selector such that if there is a proton which is having mass of 1 .67 times 10 raise to minus 27 kilograms and the charge carried is 1 .6 into 10 raise to minus 19 column.
00:28
That's having an energy equivalent to one kilo electron volt and the value of one kilo electron volt in terms of juice that's 1 .6 into 10 raised to minus 16 jules so this is the conversion factor so we have to determine the magnetic field within this velocity selector provided that the electric field magnitude that is 80000 which is 80 times 10 raise to 3 volts per meter so here we directly have the expression which relates the velocity, electric field, and the magnetic field.
01:07
But here, since we are given the energy of the proton, in that case, let's use k is equal to half mv square because the energy of this proton, it will be the kinetic energy because in the velocity selector, it's having certain speed.
01:22
So when we put the values of the energy and mass in this expression, we're going to get 1 .6 into 10 raise to minus 16 is equal to half times the mass of the proton, which is 1 .6 times 10 raise to minus 27 approximately times v square.
01:41
So from here we're going to get the value of the speed and that will be two times 1 .6.
01:49
So let's approximate this.
01:52
So we're going to cancel this and it will be two times 10 raise to minus 16 over 10 raise to minus 26...