Physics, asked by khan4827, 9 months ago

A proton of magnetic field 0.2 T enters a magnetic field perpendicular with a velocity = 6.0 × 105 m/sec. Calculate the acceleration of the proton and radius of the path.
Solution:
Given,
Magnetic field B = 0.2T
Speed of proton v = 6.0 × 105 m/s
Charge q = e = 1.6 × 10-19 C
θ = 90° ‘
∴ F = qvBsinθ
F = 1.6 × 10-19 × 6.0 × 105 × 0.2 × sin90°
F = 1.92 × 10-14 N
∵ F = ma

Answers

Answered by kyogender768
0

Answer:

plz check image bro for answer

Hope it helps ...

Don't forget to say thanks

and plz mark as brainliest answer

Similar questions