A proton in magnetic field of 2500 Newton/Ampere metre with velocity 4×10 ki power 5metre/second perpendicularly. Calculate the force acting on proton
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Answer:
A proton travels with a speed of 4.90*10^6 m/s at an angle of 62 degrees with the direction of a magnetic field of magnitude 0.150…
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2
Answer:
1.6*10 ki power-4
Explanation:
f=qvb
charge on one proton 1.6*10^-10
b= 2500 n ( a m)
F = 1.602*4*2500*10^-14
= 1.6*10^_10 n
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