Question

A point charge Q moves on the x-axis in the positive direction with a speed of A point P is on the y-axis at The magnetic field produced at point P, as the charge moves through the origin, is equal to When the charge is at what is the magnitude of the magnetic field at point P? (?0 = 4? × 10-7 T ? m/A)

Answer 1

i think values are missing

y = 60 mm = 0.06 m
Let's use a negative charge
and Fleming's Left hand rule
for the charge and the current.

For the charge as it passes the origin:
B = k*q*v / y^2 = -0.6*10^-6 k T
Note: k,q, and v are not needed.
B points along the negative z-axis at y

Now apply Ampere's law at the origin
for the current:

?c B . dL = uo I
B*2*pi*y = uo I
Solve for I
I = B*2*pi*y / uo

I = (0.6*10^-6)*2*3.14159*0.06 /
(4*3.14159*10^-7) = 0.18 A
OR
I = 180 mA, positive --->+x