Magnetic Fields Produced by Currents

[bobbworm42]

A long, straight wire is oriented in the north-south direction, and the current in the wire is directed to the north. The horizontal component of the Earth's magnetic field is 4.5 x 10-5T and points due north. A small horizontal compass needle is located directly below the wire, 1.9 cm from it. The compass needle points 40° north of west. What is the current in the wire?

So I know that:

B(horz.) = 4.5 x 10-5 T
r = .019 m
Theta = 40° North of West

I have tried combing F = ILB sin theta and B= (U(0) I)/(2*p* r), but I can find no place to incorporate my theta or how to find the vertical component of the Earth's magnetic field (if it is even needed).

Please show me the error of my ways.

 

[Mindscrape]

You may be over complicating the problem. At the bottom of the wire the magnetic field is pointing in the west direction with some magnitude based on how far away it is and how much current flows (you only know one of these). The Earth's field is pointing in the north direction. It's a vector problem.

 

[Moetasim]

I would first like to clarify that the horizontal component of Earth's magnetic field is typically denoted as B_E and the current in the wire as I. Additionally, the angle theta in this scenario would be measured from the north-south direction, not west. With that in mind, let's proceed with solving for the current in the wire.

First, we can use the formula B = (u_0 * I)/(2 * pi * r) to find the magnitude of the current in the wire. Rearranging the equation, we get:

I = (2 * pi * r * B_E)/u_0

Substituting in the given values, we get:

I = (2 * pi * 0.019 m * 4.5 x 10^-5 T)/ (4 * pi * 10^-7 T*m/A)

Simplifying, we get:

I = 0.342 A

Next, we can use the formula F = I * L * B * sin(theta) to find the force acting on the compass needle due to the magnetic field produced by the current. Since the compass needle is located directly below the wire, the angle theta would be 90 degrees. So the equation becomes:

F = I * L * B_E

Substituting in the values, we get:

F = 0.342 A * 0.019 m * 4.5 x 10^-5 T

Simplifying, we get:

F = 3.09 x 10^-7 N

This is the force acting on the compass needle due to the magnetic field produced by the current in the wire. However, we know that the compass needle is also affected by the Earth's magnetic field, which is 4.5 x 10^-5 T in the horizontal direction. This means that the compass needle is experiencing a net force of 3.09 x 10^-7 N in a direction 40 degrees north of west.

To determine the direction and magnitude of the force due to the current alone, we can use vector addition. The force due to the Earth's magnetic field can be represented as a vector pointing due north, while the force due to the current can be represented as a vector pointing 40 degrees north of west. Using trigonometry, we can find the magnitude of the force due to the current alone to be approximately 2.37 x 10^-