Question

An electron is rotating in a circular path of radius 0.01 m with frequency 10^9 Hz. Calculate the magnitude
of force experienced a proton which is moving through the center of circular path particle to the plane of
path with speed v = 1.76 x 10 m/sec
A) 0 N
B)6 N
С)16N
D)14N


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Answer 1

Answer:

the magnitude of the force experienced a portion which is moving through the centre is 16N

Answer 2

Answer:

Force experienced by proton $F=0.028*10^{-27} N$

Explanation:

Radius of the electron =0.01m

Force experienced by the proton is given by $F=qvB$

q is the charge on proton = $1.6*10^{-19}C$

v, is the velocity of the particle =$1.76*10 m/s$

we have to find B, which can be found from$\frac{mv^2}{r}=qvB$

⇒ $B=\frac{mv}{qr} =\frac{9.1*10^{-31}*1.76*10}{1.6*10^{-19}*0.01}=1.001*10^{-8}$

Putting the values of above in force

F=$1.6*10^{-19}*1.76*10*1.001*10^{-8}=0.028*10^{-27} N$

NOTE: Answers does not match any option, probably because some values given in the question is wrong.