At a certain instant of time a particle with charge q = 15 μC is located at x = 2.0 m, y = 5.0 m; its velocity at that time is v = 40 m/s . If you are at the origin, what do you measure as the magnitude of the magnetic field due to this moving point charge?

Explanation:This question uses the Biot-Savart law: the law is an equation that describes the magnetic field created by a current-carrying wire, and allows you to calculate its strength at various points.

B=(μ0/4*π)*q*v*r(unit vector)/r²

Also:

B=(μ0/4*π)*q*v*sin(θ)/r²

Where;

μ0 =permeability of free space = 4πx10-7 Hm-1

B = magnetic field in Tesla

V= velocity

r= radius

Therefore:

B=(4πx10-7/4*π)*q*v*sin(θ)/r^2

B=1x10-7*q*v*sin(θ)/r^2

Using:

q=15x10-3C

v=40m/s

tan(θ)=5/2 so θ=68.2°

r²=5²+2² (Pythagoras Theorem)

B can be calculated as:

B=1x10-7*15x10-3*40*sin(68.2)/(5²+2²)

B=1.92nT

ajeigbeibraheem ajeigbeibraheem · Answer 2 · 2020-03-28T04:02:19+01:00

Answer:

2.7 PT

Explanation:

Given that;

Charge (q) = 15 μC = [tex]15*10^{-6} C[/tex]

Velocity (V) = - 40 m/s

Position vector = 2 m î + 5 m î

Magnitude of the position vector |r| = [tex]\sqrt{(2^2)+(5^2)}[/tex] = 5.385

Angle Cos θ = [tex]\frac{v r}{|v||r|} = \frac{-40*(2i + 5j)}{40*5.385}[/tex]

= [tex]\frac{-80}{215.4}[/tex]

Cos θ = - 0.3714

θ = Cos ⁻¹ (0.3714)

θ = 111.80°

Magnitude field (B) at the origin

B = [tex]\frac{\mu q V sin \theta}{4 \pi r^2}[/tex]

B = [tex]\frac{(4 \pi *10^{-7}*15*10^{-6}*40*sin(111.80^0)}{4 \pi *(5.385)^2}[/tex]

B = [tex]2.685 *10^{-12}T[/tex]

B = 2.7 PT