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## Homework Statement

There is a wire (current I = 1.05 A) that splits into two channels; of resistance R2 = 6.60 and R1 = 1.25 , and re-joins, forming a current in the shape of an isosceles triangle with base distance d = 6.90 cm and height L = 14.0 cm. The loop is entered into the space between the two poles of a magnet with a uniform magnetic field, B = 4.55e−2 T, that runs from one pole to the other. The loop is placed such that the field lies in the plane of the loop. What is the magnitude of the torque on the circuit about the wire's axis?

(Edit: Image inserted by moderator)

## Homework Equations

F=I*

**l**x

**B**

Torque=

**r**x

**F**

3. The Attempt at a Solution

3. The Attempt at a Solution

You can picture the triangle with R2 on the left side and R1 on the right side and the wire comes, current from top to bottom right in the middle and the sides meet back up halfway at the base of the triangle.

The first thing I did was to find the current that is in each side of the wire. Since voltage drop is the same across both sides and both the currents add up to the current going in I derived a system of equation. I also tried to find the voltage using the total current and equivalent resistance and then find the individual currents using the total voltage which is the same for both of them.

I got .1672 A for the left side and .8828 A for the right side.

Now for the Force on each wire, since the current is going the same direction in each wire but they are on opposite sides of the axis of the wire the torques will be opposing directions but different strengths because of the difference currents, so using F=I*l*B*sin(theta)

I found L using pythagoreas theorem side=0.1442 m and then theta using law of cosines theta=arccos(d/2/0.1442)=76.2 deg

So l*sin(theta) would give you an arm perpendicular to the B field

F1=I1*l*sin(theta)*B and F2=I1*l*sin(theta)*B

both resulting in the direction out of the page

Now for the torque, using the center of the triangle right on the axis of the wire as reference r points straight out a distance d/3. Since the force is directed straight out of the page for both wires but the radius is in the opposite directions this gives the opposing torques, and they are perpendicular so the cross product can go away with sin(pi/2)=1

torque=I1*l*sin(theta)*B*d/3-I1*l*sin(theta)*B=|I1-I2|*l*B*sin(theta)*d/3

I ended up numerically with 1.049e-4 N*m not quite the right answer. Is this the best way of going about this problem?