Physics, asked by Nisha01, 1 year ago

A proton enters a magnetic field of intensity 10 T with a velocity of 107 m/s in a direction at an angle of 90° with the field. Force on proton is
Choose one:


3.2 × 10–11 N


1.6 × 10–11 N


1.6 × 10–19 N


5 × 10–15 N

Answers

Answered by Anonymous
7
Hi
The magnitude of the magnetic force on a charg particle is F = |q| v B sin θ
Here q = charge on a proton = 1.602 x 10^-19 C
v = 10^7m/s
θ = 90
B = 10T
Substituting all these values in the expression for F, we get
F = 1.602 x 10^-19C x 10^7 m/s x 10 T sin 90°
= 1.6 x10^-11 N

c option

hope it help
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