Question

a proton enters a magnetic field of flux density 4 t with a velocity 2.5 x 10^6 m/s at an 30 degree with the direction of the field . find the magnitude of the force acting on the proton. plz.. answer the question ....​

Answer 1

Answer:

A proton enters a magnetic field of 4 T intensity with a velocity of 2.5 × 10⁶ ms⁻¹ an angle of 30° with the field. Find the magnitude of the force on the proton . Charge on the proton = 1.602 × 10⁻¹⁹ C.

Solution:Solution:

Given :

B = 4 T

v = 2.5 × 10⁶ ms⁻¹.

We know that ,

F = q v B Sin ∅

= 1.6 × 10⁻¹⁹ × 2.5 × 10⁶ × 4 × \frac{1}{2}

2

1

simply it we get,

F = 8 × 10⁻¹³ N.