Magnitude of the magnetic field

[rayhan619]

Homework Statement

The two insulated wires in the figure (Intro 1 figure) cross at a 30^\circ angle but do not make electrical contact. Each wire carries a 5.0 {\rm A} current. Points 1 and 2 are each 3.5 cm from the intersection and equally distant from both wires.
a) What is the magnitude of the magnetic field at point 1?
b) What is the magnitude of the magnetic field at point 2?

Homework Equations

The Attempt at a Solution

 

[jbsteele]

a) B = \(\frac{\mu_0*I}{2*\pi*r}\) = \(\frac{\mu_0*5.0}{2*\pi*3.5}\) = 0.345 {\rm T} b) same answer as a)

 

[nlightner]

I would first clarify any assumptions and variables in the problem to ensure accurate calculations. In this case, I would ask for the orientation of the wires and the direction of the current flow. I would also assume that the wires are infinitely long and straight, and that the magnetic field can be calculated using the right-hand rule.

To solve for the magnitude of the magnetic field at points 1 and 2, I would use the equation B = μ₀I/(2πr), where μ₀ is the permeability constant, I is the current, and r is the distance from the wire.

a) At point 1, the distance from each wire is 3.5 cm, so the total distance is 7 cm. Using the right-hand rule, I determine that the magnetic field from wire 1 is directed out of the page and the magnetic field from wire 2 is directed into the page. The two fields will cancel each other out, resulting in a net magnetic field of 0 at point 1.

b) At point 2, the distance from each wire is also 3.5 cm. Using the right-hand rule, I determine that the magnetic field from wire 1 is directed into the page and the magnetic field from wire 2 is directed out of the page. These two fields will add together, resulting in a net magnetic field at point 2 of B = μ₀I/(2πr) = (4π×10^-7 T·m/A)(5.0 A)/(2π(0.035 m)) = 3.57×10^-5 T.

In conclusion, the magnitude of the magnetic field at point 1 is 0 and the magnitude of the magnetic field at point 2 is 3.57×10^-5 T.