Question

A Proton enters a magnetic field 2 T with a velocity of 3.4 x 10⁷ m/s the acceleration of the proton​

Answer 1

Answer:

The magnitude of the magnetic force on a charge particle F=∣q∣vBsinθ

Here, q=charge on a proton =1.602×10

−19

C

V=2×10

7

m/s

θ=30

0

B=1.5T

Substituting all these values in the expression for F

We get,

F=1.602×10

−19

C×2×10

7

m/s×1.5Tsin30

0

=2.403×10

−12

N