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 the point P, as the charge moves through the origin, is equal to What is the charge Q?

q = -52.5 μC

Explanation:

The complete question is given thus;

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 -0.30uTk. What is the charge Q? (uo=4pi x 10^-7 T m/A).

SOLVING:

from the given parameters we can solve this problem.

Given that the

Speed = 280 m/s

y = 70mm

B =  -30 * 10⁻⁶T

Using the equation for magnetic field we have;

Β = μqv*r / 4πr²

making q (charge) the subject of formula we have that;

q  = B * 4 *πr² / μqv*r

substituting the values gives us

q = (-0.3*10⁻⁶Tk * 4π * 0.07²) / (4π*10⁻⁷ * 280 ) = - [14.7 * 10⁻¹⁰k / 2.8 * 10⁻⁵ k ]

q = -52.5 μC

cheers i hope this helped !!!

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