An electron moves in a circular path perpendicular to a constant magnetic field with a magnitude of 2.00 mT. The angular momentum of the electron about the center of the circle is 3.50 *10^-25 J·s.

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An electron moves in a circular path perpendicular to a constant magnetic field with a magnitude of 2.00 mT. The angular momentum of the electron about the center of the circle is 3.50 multiplied by 10-25 J·s.

(a) Determine the radius of the circular path.
 cm
(b) Determine the speed of the electron.
 m/s
asked Feb 10, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
    

1 Answer

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Best answer

Start by relating magnetic force to force:

magnetic force = q*v*B (*where q=charge, v=velocity, B=magnetic field)

force = mass*acceleration = (m*v^2) / r  (*where m=mass, r=radius)

set the two equations equal and solve for the velocity:

q*v*B = (m*v^2) / r

solving for v we get:

v = (q*B*r) / m



We are given angular moment, L, so lets write out the equation for L.

L = m*v*r

Subsituting v from above into the equation we get:

L = m*((q*B*r) / m)*r  =  q*B*r ^2

Now we can solve this equation for r:

 

r = squareroot (L / (q*B))

where L = 3.50 multiplied by 10-25 J·s.

magnetic field (B) = 2.00 mT = 2*10^-3 T

q = 1.602*10^-19 C

 

plugging these values in we get:

r = squareroot ((3.50 multiplied by 10-25 J·s.) / ((1.602*10^-19 C) * (2*10^-3 T)))

r = .03305 m = 3.305 cm

 

Part B:

In Part A, we found that velocity is:

v = (q*B*r) / m

and now that we have r, we can plug that in along with the other variables and find v:

(mass of electron = 9.109*10^-31 kg)

v = ((1.602*10^-19 C)*(2*10^-3 T)*(.03305 m)) / (9.109*10^-31 kg)

v = 11625881 m/s

answered Feb 19, 2012 by pokemonmaster ~Expert~ (3,856 points)
selected Feb 19, 2012 by Joey33
does the electron in circular path radiate?

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