Current in a rectangle on a hinge

[lodovico]

Homework Statement

Figure 29-36 shows a rectangular, 15-turn coil of wire, 10 cm by 5.0 cm. It carries a current of 0.90 A and is hinged along one long side. It is mounted in the xy plane, at an angle of 30° to the direction of a uniform magnetic field of 0.50 T. Find the magnitude and direction of the torque acting on the coil about the hinge line.

Homework Equations

$\tau$=μ × B
μ=NiA

The Attempt at a Solution

I don't know how to approach this. I tried to plug in numbers into ^ formula

 

[TSny]

If you'll show more details of how you tried the calculation, we will be able to see where you're having trouble.

 

[lodovico]

τ=μ × B
μ=NiA

τ=NiABsin30
τ=(15)(.9)(.1*.05)(.5)sin30

τ=.16875 Nm

 

[TSny]

τ=μ × B
μ=NiA

τ=NiABsin30
τ=(15)(.9)(.1*.05)(.5)sin30

τ=.16875 Nm

See if you can figure out why the 30^o is not correct here. You'll need to think about the direction of $\vec{\mu}$

 

[Nickie]

but I don't think I'm on the right track.
I would approach this problem by first identifying the key variables and equations that are relevant to the situation. In this case, we have a rectangular coil of wire carrying a current in a magnetic field, and we are trying to find the torque acting on the coil.

The key equations that are relevant here are the torque equation, which is given by τ=μ×B, and the equation for magnetic moment, which is given by μ=NiA.

To solve this problem, we need to first calculate the magnetic moment of the coil. We can do this by multiplying the number of turns (N) by the current (i) and the area of the coil (A). In this case, N=15, i=0.90 A, and A=10 cm x 5.0 cm = 50 cm^2. Converting to SI units, we get μ = (15)(0.90 A)(0.005 m^2) = 0.0675 A·m^2.

Next, we can calculate the torque by taking the cross product of the magnetic moment and the magnetic field. The direction of the torque will be perpendicular to both the magnetic moment and the magnetic field, according to the right-hand rule. In this case, the angle between the magnetic moment and the magnetic field is 30°, so we can use the formula τ = μBsinθ. Plugging in our values, we get τ = (0.0675 A·m^2)(0.50 T)(sin 30°) = 0.0338 N·m.

Therefore, the magnitude of the torque acting on the coil is 0.0338 N·m, and the direction is perpendicular to both the magnetic moment and the magnetic field, in the direction determined by the right-hand rule.

In conclusion, by using the relevant equations and plugging in the given values, we were able to calculate the magnitude and direction of the torque acting on the coil about the hinge line. This approach can be applied to similar problems involving magnetic fields and electric currents.