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

Question

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Respuesta :

Xema8 Xema8 · Answer 1 · 2019-11-04T12:37:07+01:00

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 = [tex]\frac{10^{-7}\times108\times10^{-4}\times60}{72\times72\times72\times10^{-6}}[/tex]

= 0.17 μT

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-31T14:04:45+02:00

t=The magnitude of the magnetic field at point P is mathematically given as

B= 0.17 uT

Question Parameter(s):

A point charge Q moves on the x-axis in the positive direction with a speed of 290 m/s

the y-axis at y = +60 mm

is equal to -0.3 μT k^.

x = +40 mm

Generally, the equation for the Magnetic field is mathematically given as

B = (u / 4\pi) x ((qv x r)/r^2)

Therefore

0.3 x 10^{-6} = 10^{-7} x qvsinθ /( 60 x 10^{-3})^2

qv = 10.8 x 10^{-3}

In conclusion

[tex]B = \frac{10^{-7}*108*10^{-4}*60}{72*72*72*10^{-6}}[/tex]

B= 0.17 uT