A very long, thin rod carries electric charge with the linear density 34.0 nC/m.
ID: 1864717 • Letter: A
Question
A very long, thin rod carries electric charge with the linear density 34.0 nC/m. It lies along the x axis and moves in the x direction at a speed of 15.0 Mm/s (a) Find the electric field the rod creates at the point (0, 18.0 cm, 0) Magnitude N/CDirection -z axis -y axis +z axis ty axis (b) Find the magnetic field it creates at the same point. Magnitude TDirection ty axis y axis -z axis +z axis (c) Find the force exerted on an electron at this point, moving with a velocity of 240 i Mm/s Magnitude NDirection ty axis y axis +z axisExplanation / Answer
(a) Magnitude: E=lambda/(2*pi*epsilon*r)
where lambda is the linear density, epsilon is permittivity of free space (8.854*10^-12) r is the radius (ie. 18cm)
= (34*10^-9)/(2pi*8.854*10^-12 *0.18)
= 3395 N/C
(b) (b) because the rod is moving with velocity +15.0 Mm/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 = 34.0 nC/m * 15.0 Mm/s = 34.0 *15.0 * 10^-9 * 10^6 C/m
I = 0.51 Amperes
B = mu * I / (2*pi*r)
I is above, r = 18.0 cm for the given point
So
B = (4pi*10^-7 * 0.51)/(2pi*0.18)
= 5.67 x 10^-7 T
(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*10^6 * B
= (1.6*10^-19)( 240*10^6) * (5.67 x 10^-7)
= 2.176 x 10^-17 N
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