Question

A proton enters in a magnetic field of flux density 2.5T with a velocity of 1.5x107ms-1 at an angle of

30degree with the field find the force on the proton​

Answer 1

Answer:

3×10^-12 N

Explanation:

we know that force on a charge due to magnetic field is given as:

Fm=qvBsin¢

¢=30°

q=1.6×10^-¹⁹C

B=2.5T

v=1.5×10^⁷

thus,Fm=1.6×10^-¹⁹×1.5×10^⁷×2.5×sin30

Fm=3×10^-¹² N. (sin30=1/2)