Question

Proton moving with V=2*106 m/s speed in a magnetic field of B=3T is under the effect of 4*10-12 N force. What is the angle between the magnetic field and the velocity of the proton.

Answer 1

Note :- In the given problem I think proton velocity must be 2×10^7m/s instead of given value because if we take into account, getting math error so it has changed with power 7

Given data :-

charge of a proton (q) = + 1.602×10^-19 C

Velocity of proton enters with speed (V)= 2×10^7 m/s

Magnetic field (B) = 3 T

Force (F) = 4×10^-12 N

Expression for the force in a magnetic field :-

F = qVB*sin

sin = F/(qVB) ---- Equation

subsitute the given values in above equation ,

We get ;

sin = 4×10^-12 / (1.602×10^-19*2×10^7* 3)

sin = 4×10^-12 / 9.612×10^-12

sin = 0.4161

= (0.4161)

= 24.59 °

The angle () between magnetic field (B) and Velocity (V) is   = 24.59 deg

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