Question

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?

Answer 1

The question is missing. Here is the complete question.

A point charge Q moves on the x-axis in the positive direction with a speed of 160 m/s. A point P is on the y-axis at y = + 20mm. The magnetic field produced at point P, as the charge moves through the origin is equal to -0.6μT k. What is the charge Q?
`\mu_(0)=4.\pi.10^(-7)`T.m/A)

Answer: Q =
`-0.015.10^(-3)`C

Step-by-step explanation: Magnetic Field (B) is a vector field, i.e., has magnitude and direction, and describes the distribution of magnetic force in the space around. It happens when an electrical charge is in movement.

Its magnitude is determined by the formula:

`B = (\mu_(0))/(4.\pi)(Qvsin(\theta))/(r^(2))`

where

`\mu_(0)` is the vacuum permeability constant;

r is the distance between charge and a point you want to know the magnetic field;

θ is the angle between velocity and distance r;

For this question, magnetic field has that magnitude when charge is passing through the origin. So, angle between velocity and distance is 90°.

Calculating to determine charge:

`-0.6.10^(-6) = (4.\pi.10^(-7))/(4.\pi)(Q.160.sin(90))/((2.10^(-2))^(2))`

`-0.6.10^(-6) = 10^(-7)(Q.160.1)/((2.10^(-2))^(2))`

`-2.4.10^(-10) = Q.160.10^(-7)`

`Q = (-2.4.10^(-10))/(160.10^(-7))`

`Q = -0.015.10^(-3)`

Charge Q is
`-0.015.10^(-3)`C or
`-0.015`mC.