A very long, thin rod carries electric charge with the linear density 28.0 nC/m.

Question

A very long, thin rod carries electric charge with the linear density 28.0 nC/m. It lies along the x axis and moves in the x direction at a speed of 15.0 Mm/s.

A very long, thin rod carries electric charge with the linear density 28.0 nC/m. It lies along the x axis and moves in the x direction at a speed of 15.0 Mm/s. Find the electric field the rod creates at the point (0, 24.0 cm, 0). Magnitude: 2098 N/C Find the magnetic field it creates at the same point. Magnitude: 3.5 e-7 T Find the force exerted on an electron at this point, moving with a velocity of 240 Mm/s.

Explanation / Answer

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) because the rod is moving with velocity +15.0 m/s, a B field is created from the moving charge. So need to compute the current:
I = dq/dt

Look at the charge that passes through the plane perpendicular to the x-axis (y-z-plane).
So start counting charge at time t=0 until time t=dt,
in that time dit the rod has moved ds = v * dt = 15.0 * dt, so a total charge of
lambda * ds has past through the y-z plane.

I = lamda * ds/dt = lambda * v = 36.0 nC/m * 15.0 m/s = 36.0 *15.0 * 10^-9 C/m
I = 36.0 *15.0 * 10^-9 Amperes

B = mu * I / (2*pi*r)
I is above, r = 16.0 cm for the give point

Then compute B with the appropriate units.
(c)
F = qv x B_vec, were B_vec has a direction out of the plane (use right hand rule with thumb pointed in the +x direction.

v in direction i (+x), so it is perpendicular since B must be in the +z direction out of the page.
v x B = v*B * sin(theta_vB) = v*B*sin(90) = v*b

F = e * 240 * B

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