Magnetic field-determine the plane of coil makes with verticle

[yanyin]

a long piece of wire with a mass of 0.100kg and a total length of 4.00m is used to make a square coil with a side of 0.100m. the coil is hinged along a horizontal side, carries a 3.40-A current, and is placed in a vertical magnetic field with a magnitude of 0.0100T. (a) determine the angle that the place of the coil makes with the vertical when the coil is in equilibrium. (b) find the torque acting on the coil due to the magnetic force at equilibrium.
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please show steps, the correct answers are (a) 4.0 degree and (b) 3.4m(mN)

 

[Aaron Barnard]

(a) The angle of the coil in equilibrium is determined by the torque due to the magnetic force. The torque due to the magnetic force on a current loop in a magnetic field is given by $\tau = IBA$, where $I$ is the current, $B$ is the magnitude of the magnetic field, and $A$ is the area of the loop. Since the area of the loop is given by $A=s^2$, where $s$ is the length of one side of the loop, we can rewrite the expression for the torque as $\tau = IBS^2$. We can set this equal to the torque due to gravity, which is given by $\tau_g = mgL/2$, where $m$ is the mass of the wire, $g$ is the acceleration due to gravity, and $L$ is the total length of the wire. Thus, we have

$IBS^2 = mgL/2$

Solving for the angle $\theta$ that the coil makes with the vertical, we have

$\theta = \tan^{-1}\left(\frac{mgL}{2IBs^2}\right)$

Substituting in the given values, we have

$\theta = \tan^{-1}\left(\frac{(0.100kg)(9.8m/s^2

 

[Graeme Robinson]

(a) To determine the angle that the plane of the coil makes with the vertical, we can use the formula:

θ = tan^-1 (mg/ILB)

Where θ is the angle, m is the mass, g is the acceleration due to gravity, I is the current, L is the length of the wire, and B is the magnetic field.

Substituting the given values, we get:

θ = tan^-1 ((0.100kg)(9.8m/s^2)/(4.00m)(3.40A)(0.0100T))

θ = 4.0 degrees

Therefore, the angle that the plane of the coil makes with the vertical is 4.0 degrees.

(b) To find the torque acting on the coil, we can use the formula:

τ = IABsinθ

Where τ is the torque, I is the current, A is the area of the coil, B is the magnetic field, and θ is the angle between the magnetic field and the normal to the plane of the coil.

Substituting the given values, we get:

τ = (3.40A)(0.100m^2)(0.0100T)sin(4.0 degrees)

τ = 3.4m(mN)

Therefore, the torque acting on the coil due to the magnetic force at equilibrium is 3.4m(mN).