Question

A point charge Q moves on the x-axis in the positive direction with a speed of 370 m/s. A point P is on the y-axis at y = + 80 mm. The magnetic field produced at point P, as the charge moves through the origin, is equal to - 0.8 μT k^. When the charge is at x = + 40 mm, what is the magnitude of the magnetic field at point P? (μ0 = 4π × 10-7 T • m/A)

A. 0.57 μT
B. 1.1 μT
C. 0.92 μT
D. 0.74 μT

Answer 1

To find the magnitude of the magnetic field at point P, we can use the formula for the magnetic field produced by a moving charge. By plugging in the values and solving for the charge, we find that the magnitude of the magnetic field at point P is 0.57 μT.

Step-by-step explanation:

To find the magnitude of the magnetic field at point P, we can use the formula for the magnetic field produced by a moving charge: B = (μ0Qv)/(4πr^2). In this formula, B is the magnetic field, μ0 is the permeability of free space, Q is the charge, v is the velocity, and r is the distance between the charge and the point where the magnetic field is being measured. Since the charge is moving on the x-axis and the point P is on the y-axis, the distance between them is the y-coordinate of P, which is +80 mm.

Plugging in the values into the formula, we have B = (4π × 10-7 T • m/A)(Q)(370 m/s)/(4π(80 × 10-3 m)2). The 4π cancels out, so we're left with B = (10-7)(Q)(370 m/s)/(80 × 10-3)2. Since we know that the magnetic field at point P is -0.8 μT k^, we can equate it to the calculated magnetic field and solve for Q.

-0.8 μT k^ = (10-7)(Q)(370 m/s)/(80 × 10-3)2. Solving for Q, we find that Q ≈ -8 μC.