A very long, thin rod carries electric charge with the linear density 30.0 nC/m. It lies along the x axis and moves in the xdirection at a speed of 15.0 Mm/s.

+7 votes

A very long, thin rod carries electric charge with the linear density 30.0 nC/m. It lies along the x axis and moves in the xdirection at a speed of 15.0 Mm/s.

(a) Find the electric field the rod creates at the point (0, 24.0 cm, 0). 
Magnitude: 
 N/C
Direction: 

(b) Find the magnetic field it creates at the same point. 
Magnitude: 
 T
Direction: 

(c) Find the force exerted on an electron at this point, moving with a velocity of 240 ihatbold Mm/s. 
Magnitude: 
 N
Direction:
asked Mar 27, 2012 in Physics by yoshi ~Expert~ (919 points)
    

2 Answers

+3 votes
I only figured out part a) answer. hopefully someone else can get b) or c)

a) use the equation E=lambda/(2*pi*epsilon*r)

where lambda is the linear density, epsilon is permativity of free space (8.854*10^-12) r is the radius (ie. 24cm)

*note: convert units of nC to C and cm to m

it's in +y direction

b) it's in +z direction

c) it's in -y direction
answered Apr 1, 2012 by jarjar123 ~Rookie~ (131 points)
edited Apr 1, 2012 by jarjar123
I will post parts  B and C
+3 votes

Part B:

The change in motion constitutes a current:

Current = wavlength * velocity

Plugging this current into the formula for the magnetic field we get:

(where * B=magnetic field, u=4pi*10^-7 , I=current, a=radius)

 

Part C:

Lorentz Force is equal to:

F = qE + qv X B   (*where F=force q=charge, E=electric field, v=velocity, B=magnetic field, and X=cross product)

 

answered Apr 1, 2012 by Roar3215 ~Expert~ (1,318 points)
edited Apr 1, 2012 by Roar3215

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