Question

He^(2+) ion travels at right angles to a magnetic field of 0.80 T with a velocity of 10^5 m/s. Find the magnitude of the magnetic force on the ion.

Answer 1

Magnitude of magnetic force on the ion = 2.56 * 10^-14 N

->   Given: Charge on ion = 2 * 1.6 * 10^-19 C; B = 0.8 T; v = 10^5 m/s;

->   Magnetic Force (F) = Bqv = 0.8 * 2 * 1.6 * 10^-19 * 10^5 = 2.56 * 10^-14 N

Answer 2

The magnitude of magnetic force is determined by the equation:-

F=qvB

where q= charge on He^2+ = 2* 1.6*10^-19

           v= velocity of the particle

           B= magnetic field strength

So, F= 2 * 1.6 * 10^-19  * 10^5*0.8*

     F=  2.56 * 10^-14 N

Here the force exerted by the He^2+ is 2.56 * 10^-14 N