Respuesta :
Complete Question
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 ? (uo=4pi x 10^-7 T m/A)
Answer:
The value is [tex]B = -0.1485 \mu \ T[/tex]
Explanation:
From the question we are told that
The speed of the charge is [tex]v = 280 \ m/s[/tex]
The position of point P from the origin on the y-axis is [tex]y = 70 \ mm = 70*10^{-3} \ m[/tex]
The magnitude of the magnetic field when the charge is at origin is [tex]B_o = -0.3 \ \mu T = -0.3 *10^{-6} \ T[/tex]
The position of the charge considered is [tex]x = 70 \ mm = 70 *10^{-3}[/tex]
Generally the magnitude of the charge is mathematically represented as
[tex]q= \frac{B_o * (4 \pi * y ^2 )}{\mu_o * v }[/tex]
=> [tex]q= \frac{ ( -0.3 *10^{-6})* (4 * 3.142 * (70*10^{-3}) ^2 )}{ 4 * 3.142 * 10^{-7} * 280 }[/tex]
=> [tex]q= -52 *10^{-6} \ C[/tex]
Here the negative sign show that the charge is negative signed
Generally applying Pythagoras theorem , the distance from the considered position of the charge to point P is mathematically represented as
[tex]r = \sqrt{ y^2 + x^2 }[/tex]
=> [tex]r = \sqrt{ (70*10^{-3})^2 + (70*10^{-3})^2 }[/tex]
=> [tex]r = 0.09899 \ m[/tex]
Gnerally the magnitude of the magnetic field at point P when the charge is at x is mathematically represented as
[tex]B = \frac{\mu_o * q * v }{ 4 \pi * r^2 }[/tex]
=> [tex]B = \frac{4\pi * 10^{-7} * -52 *10^{-6} * 280 }{ 4 \pi * 0.09899^2 }[/tex]
=> [tex]B = \frac{4 * 3.142 * 10^{-7} * -52 *10^{-6} * 280 }{ 4* 3.142 * 0.09899^2 }[/tex]
=> [tex]B = -0.1485 *10^{-6 } \ T[/tex]
=> [tex]B = -0.1485 \mu \ T[/tex]
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; https://brainly.com/question/13720291
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)
