Question

A point charge Q moves on the x-axis in the positive direction with a speed of 290 m/s. A point P is on the y-axis at y = +60 mm. The magnetic field produced at point P, as the charge moves through the origin, is equal to -0.3 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)
0.22 μT
0.38 μT
0.17 μT
0.33 μT
0.28 μT

Answer 1

Answer:

Explanation:

Magnetic field due to a moving charge

B = μ₀ / 4π x (qv x r)/r²

Given

B = .3 X 10⁻⁶ T,

r = 60 x 10⁻³ m

Putting these values in the given equation

.3 x 10⁻⁶ = 10⁻⁷ x qvsinθ /( 60 x 10⁻³)²

θ = 90 °

qv = 10.8 x 10⁻³

At point P

Sinθ  = 60 / 72

r² = 72 x 72 x 10⁻⁶

B = $\frac{10^{-7}\times108\times10^{-4}\times60}{72\times72\times72\times10^{-6}}$

= 0.17 μT