Question

A high-speed particle is moving 8.0x107 m/s in a uniform magnetic field of 3.0 T. The...

A high-speed particle is moving 8.0x107 m/s in a uniform magnetic field of 3.0 T. The particle's mass is 4.8x10-24 kg with a charge of 6.5x10-16 C. Please show your work.

a.) Calculate the radius the particle is traveling?

b.) In electron volts calculate Kinetic Energy?

c.) What is the potential difference the particle accelerated due to its Kinetic Energy?

Homework Answers

Answer #1


apply centripetal force = magnetic force

i.e mv^2/r = qvB,

where m = mass o the charged particle

v = velocity,   r = radius ,

q = charge = 1.6*10^-19 C

B = magnetic field

so now , so as r    = mv/qB

r = (4.8 *10^-24 * 8 *10^7)/(6.5*10^-16 * 3)

raidus r = 19.6 cm

----------------------------------

b. KE = 0.5 mv^2

KE = 0.5 * 4.8 *10^-24 * (8*10^7)^2

KE = 1.5 *10^-8 J

1 eV = 1.6 *10^-19 J

so

KE = 1.5 *10^-8/(1.6*10^-19)

KE = 9.37 *10^10 eV

-------------------------

PE = e V    =1.5 *10^-8 J

Potential Diff V = (1.5 *10^-8)/(6.5 *10^-16)

V = 23 MVolts

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