Physics, asked by mondayhunter, 8 months ago

A proton moving with a velocity 3* 10^5 enters a magnetic field of 0.3T at an angle of 30degree with the field. The radius of curvature of its path will be .
a)3Q/2
b)Q/2
c)Q/4
d)Q

Answers

Answered by Brainlyguru01
1

Answer:

3Q/2

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