Question

proton enters the magnetic field of 1.5T with the speed of 2×10^7 ms^-1 at an angle of 30 degree with the field​

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

Answer 2

Answer:

1.5t field is the answer