Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point x = 5.0 cm on the x axis. The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin?

Respuesta :

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

The magnitude of magnetic field at the origin is required.

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

x = Location at x axis = 5 cm

y = Location at y axis = 4 cm

[tex]I_x[/tex] = Current at the x axis point = 50 A

[tex]I_y[/tex] = Current at the y axis point = 80 A

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnitude of the magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

Finding the net magnetic field using the Pythagoras theorem

[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

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