00:01
Hello students in this question we are given that there is an ion with charge plus 6e so charge on ion is plus 6e which is traveling horizontally to the left at a velocity of so velocity is given to a switch 9 .00 kilometers per second when it enters the magnet field that is perpendicular to its velocity and deflects it downward with the initial magnetic force of 7 of 4 4 .94, 4 .94 multiplied by 10 to the power minus 15 newton.
00:38
So in the a part we have to find the direction of the magnetic field.
00:45
Here using right hand rule, the direction of magnetic field, the direction of magnetic field will be, will be, will, b into the page into the or we can say into the screen so according to right hand rule at index finger points in the direction of moving charge and at thumb points or and our middle finger point in the direction of the magnetic field then at thumb points in the direction of the force so using this rule we find the direction of magnetic field will will be into the screen now in the b part we have to find the magnitude of this field now we know that the force is q v b sine theta since the magnetic wheel is perpendicular to the charge so we have if is equal to qvb sine 90 degree so we have force as qvb so force now we have to find the magnetic field so from here b will be equal to f over qv.
02:20
So b is equal to forces 4 .94 multiply by 10 to the 1 minus 15 over q that is plus 6 is 6 multiple by charge 1 .1 .1 is 1 .6 multiply by 10 to the bar minus 19.
02:37
Multiply by v is 9 kilometers per second.
02:41
So it will be 9 multiplied by 10 to the power 3 and it will be in minimum.
02:44
Meters per second...