When t = 2.00 s, calculate the magnitude of the force exerted on an electron located at point P1, which is at a distance r1 = 5.70 cm from the center of the circular field region.

+5 votes

Within the green dashed circle shown in the figure below, the magnetic field changes with time according to the expression B = 7.00t3 − 2.00t2 + 0.800, where B is in teslas, t is in seconds, and R = 2.85 cm.

image
(a) When t = 2.00 s, calculate the magnitude of the force exerted on an electron located at point P1, which is at a distance r1 = 5.70 cm from the center of the circular field region.
 N

(b) When t = 2.00 s, calculate the direction of the force exerted on an electron located at point P1, which is at a distance r1 = 5.70 cm from the center of the circular field region.
    

(c) At what instant is this force equal to zero?
 s

 

asked Feb 28, 2012 in Physics by awesome ~Expert~ (1,479 points)
edited Mar 3, 2012 by awesome
    

1 Answer

+4 votes
 
Best answer

Force = E*q     (*where E=electric field, q=charge (of electron) = 1.602*10^-19 C )


To find force, first find E.

Emf = integral( E*ds) =  - dΦB/dt

B/dt = B*A   (*where A=area)

A = (pi*R^2)  

So, we can now solve for E, by taking the derivative of Emf:

integral( E*ds) = B*A       becomes...

E = dB*A/ds    

**note ds = 2pi*r1

We find dB by taking the derivative of B:

B = 7.00t3 − 2.00t2 + 0.800, 

dB = 21t2 - 4t, at time t=2, so

dB = 76

Plugging dB and A into the equation for E we get:

 

 

 
E = 76*(pi*R^2) /(2pi*r1)  
 
R = 2.85 cm = 2.85*10^-2 m
 
r1 = 5.7 cm = 5.7*10^-2 m
 
so,
 
E = 76*(pi*(2.85*10^-2 m)^2) /(2pi*5.7*10^-2 m)  
 
 
 
Now we can plug this back into the equation for Force,
 
 
 
Force = (76*(pi*(2.85*10^-2 m)^2) /(2pi*5.7*10^-2 m)  )*(1.602*10^-19 C)
 
Using Right Hand Rule, we know this force is clockwise
 
Part C,
The force is equal to zero dB/dt = 0
dB = 21t- 4t = 0
 
 
 
Solving for t, we get t = 4/21 s
answered Mar 7, 2012 by pokemonmaster ~Expert~ (3,856 points)
edited Apr 19, 2012 by pokemonmaster
why is the A=pi*R^2/2pir1
Thanks for catching this, The 2pi*r is actually the ds part of (E*ds), I had divided it over to the other side of the equation combined it with the A term, I have edited the answer to make it more clear.

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