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? (μ0 = 4π × 10-7 T • m/A)

The magnitude of the magnetic field at point P when the charge is at x = +0.07 m is; B = -0.15 uT

The formula for magnetic field due to a moving charge is given by;

B = μ₀qv/(4πr²)

Where;

μ₀ is a constant = 4π × 10⁻⁷ T.m/A

q is the charge

v is speed

r is distance of charge from point P

We are given;

B = -0.3uT = 0.3 × 10⁻⁶ T

Speed; v = 280 m/s

Position from y = 70 mm = 0.07 m

Position from x = 70 mm = 0.07 m

r = 60 x 10⁻³ m

Let us now find the value of the charge q;

From B = μ₀qv/(4πr²)we have;

q = (4πBr²)/(μ₀v)

For the value of R, we will use y = 0.07 m

Thus;

q = (4π × -0.3 × 10⁻⁶ × 0.07²)/(4π × 10⁻⁷ × 280)

Thus;

q = -5.25 × 10^(-5) C

When charge is at x = +0.07 m, it means that the distance r is the resultant of the distance in the x and y directions;

r = √(0.07² + 0.07²)

r = 0.099 m

Thus;

B = (4π × 10⁻⁷ × -5.25 × 10^(-5) × 280)/(4π × 0.099²)

B = -0.15 uT

Read more about magnetic field charge at; link

Complete question is;

A point charge Q moves on the x-axis in the positive direction with a speed of 280 m/s. A point P is on the y-axis at y=+70mm. The magnetic field produced at the point P, as the charge moves through the origin, is equal to -0.30uT . When the charge is at  x = + 70 mm what is the magnitude of the magnetic field at point P ? (μ₀ = 4π × 10⁻⁷ T.m/A)

answer; (d) spontaneous processes are characterized by the overall conversion of order to disorder;