Radius = .18m To achieve this speed v, particles are accelerated in a region wit

Question

Radius = .18m

To achieve this speed v, particles are accelerated in a region with a potential difference V (inside the electron gun, remember?). Use equations 1 and 2 obtained in this prelab to come up with an expression for the magnetic field B as a function of r, q, m, and V B= [Eq. 3] Let's test your expression. An electron originally at rest is accelerated by aV potential difference between the electrodes. The electron passes through a hole in one of the electrodes and enters a region of uniform magnetic field, with its velocity vector perpendicular to the field. The electron is observed to describe a circular orbit of radiusm. What is the magnitude of the magnetic field, in mT? (Note the units, milliteslas!!) 5. Mass of the electron: 9.1 x 101 kg Charge of the electron: 1.6 × 10-19 C

Explanation / Answer

From the given question,

charge=q

potential difference=V

mass =m

velocity=v

electrostatic potential energy = kinetic energy

q V= (1/2) mv2

v= (2qV/m)1/2

force due to magnetic field= centripetal force

Bqv= mv2/r

B=mv/(qr)

B= [m/(qr)](2qV/m)1/2

=(2mV/qr2)1/2

5. B=(2mV/qr2)1/2

=[2*9.1*10-31*4.8/(1.6*10-19*0.182)]1/2

=0.041* 10-3 T

=0.041 mT

Magnitude of magnetic field is 0.041 mT

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