Question

A proton (e = 1.60 X 10^{-19} C and m = 1.67
X 10^{-27} kg) is traveling at 90.0° with respect to the
direction of a magnetic field of strength 4.50 mT experiences a
magnetic forceof 7.50 X 10^{-17} N. The
proton's kinetic energy is:

Answer #1

Magnetic force on a moving charge particle is given by:

F = q*V*B*sin

V = speed of proton = F/(q*B*sin )

Using given values:

B = magnetic field = 4.50 mT = 4.50*10^-3 T

q = charge on ealectron

F = Magnetic force on proton = 7.50*10^-17 N

So,

V = 7.50*10^-17/(1.60*10^-19*4.50*10^-3*sin 90 deg)

V = 104166.67 m/s

Now kinetic energy of proton will be:

KE = (1/2)*m*V^2

KE = (1/2)*1.67*10^-27*104166.67^2

**KE = 9.06*10^-18 J**

If you need answer in eV, then 1 eV = 1.6*10^-19 J, So

KE = (9.06*10^-18)/(1.6*10^-19) eV = 56.6 eV (Use only if you need final answer in eV)

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