An electron at point A in (Figure 1) has a speed v 0 of 1.43×10 6 m/s . Part A)

Question

An electron at point A in (Figure 1) has a speed v0 of 1.43×106 m/s .

Part A) Find the direction of the magnetic field that will cause the electron to follow the semicircular path from A to B.

Please choose one of the following: The magnetic field will go... into the page, out of the page, left, or right

Part B) Find the magnitude of the magnetic field that will cause the electron to follow the semicircular path from A to B.

Part C) Find the time required for the electron to move from A to B.

Part D) What magnetic field would be needed if the particle were a proton instead of an electron?

Explanation / Answer

A) According to right hand rule the direction of magnetic field will be inward or into the page.

B) V0 = velocity = 1.43*106 m/s

force ,F = q*V0*B*sin(theta)

[theta is the angle between V and B which is 90 here,sin90 =1)

F= q*V0*B -------------------------------1

this force will provide the necessary centripetal force, F= m*Vo2 / R ----------------------2

from 1 and 2

q*V0*B = m*Vo2 / R ( R = radius = 5 cm(from fig)

B= m*V0 / q*R

B(magnetic field) = 9.1*10-31 *1.43*106 / 1.6*10-19 *5*10-2

B= 1.626*10-4 T

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

C) time = distance / velocity

distance = pi*R

time = pi*R / V0

time = 3.14*5*10-2 / 1.43*106 = 1.097*10-7 s

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

d) B= m*V0 / q*R

m= mass of proton = 1.67*10-27 kg

B= 1.67*10-27 *1.43*106 / 1.6*10-19 *5*10-2

B= 0.298 T

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