An eight turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm. The coil lies in the plane of the screen and has a 6.10 A current flowing clockwise around it. If the coil is in a uniform magnetic field of 2.07 10-4 T directed toward the left of the screen, what is the magnitude of the torque on the coil? (Hint: The area of an ellipse is A = πab, where a and b are, respectively, the semimajor and semiminor axes of the ellipse.)

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To find the magnitude of the torque on the coil, we can use the formula for the torque on a current-carrying coil in a magnetic field, which is:

τ = NIAωsinθ

Where:
τ = torque
N = number of turns in the coil
I = current flowing through the coil
A = area enclosed by the coil
ω = angular velocity of the coil
θ = angle between the magnetic field and the normal to the plane of the coil

First, let's calculate the area of the elliptical coil using the given major and minor axes:

a = 40.0 cm = 0.4 m (major axis)
b = 30.0 cm = 0.3 m (minor axis)

The area of an ellipse is given by A = πab. Plugging in the values, we get:

A = π(0.4 m)(0.3 m)
A = 0.12π m^2

Now, let's calculate the torque on the coil. We know that the coil has 8 turns, a current of 6.10 A flowing clockwise, and is in a uniform magnetic field of 2.07 × 10^(-4) T directed toward the left of the screen.

N = 8 (number of turns)
I = 6.10 A (current)
A = 0.12π m^2 (area)
θ = 90 degrees (since the magnetic field is perpendicular to the plane of the coil)

To convert θ from degrees to radians, we use the conversion factor π/180:

θ = 90 degrees × π/180
θ = π/2 radians

Finally, let's calculate the torque on the coil:

τ = (8)(6.10 A)(0.12π m^2)(2.07 × 10^(-4) T)(π/2)
τ = (8)(6.10)(0.12)(2.07 × 10^(-4))(π^2/2)
τ ≈ 0.022 Nm (rounded to three decimal places)

Therefore, the magnitude of the torque on the coil is approximately 0.022 Nm.