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.)

To find the magnitude of the torque on the coil, we can use the formula:

τ = NIABsinθ

where:
τ is the torque,
N is the number of turns in the coil,
I is the current flowing through the coil,
A is the area of the coil,
B is the magnetic field strength,
θ is the angle between the magnetic field and the normal to the coil.

Let's calculate step-by-step.

Step 1: Determine the number of turns in the coil.
Given: N = 8 turns

Step 2: Calculate the area of the coil.
Given: Major axis (a) = 40.0 cm = 0.4 m
Minor axis (b) = 30.0 cm = 0.3 m

Using the formula for the area of an ellipse: A = πab

A = π * 0.4 * 0.3
A ≈ 0.3769911 m^2

Step 3: Determine the angle between the magnetic field and the normal to the coil.
Given: The magnetic field is directed toward the left of the screen.

Since the coil lies in the plane of the screen, the angle between the magnetic field and the normal to the coil is 90 degrees.

Step 4: Calculate the torque.
Given: I = 6.10 A
B = 2.07 × 10^-4 T
θ = 90 degrees = π/2 radians

Using the formula: τ = NIABsinθ

τ = 8 * 6.10 * 0.3769911 * 2.07 × 10^-4 * sin(π/2)
τ = 0.00242348 Nm

Therefore, the magnitude of the torque on the coil is approximately 0.00242348 Nm.

To find the torque on the coil, we need to use the formula:

Torque = N * A * B * sin(θ)

Where:
- N is the number of turns of the coil
- A is the area of the coil
- B is the magnetic field strength
- θ is the angle between the magnetic field and the normal to the coil

First, let's find the area of the elliptical coil using the given formula:

A = π * a * b

Given:
- Major axis (2a) = 40.0 cm, so a = 20.0 cm
- Minor axis (2b) = 30.0 cm, so b = 15.0 cm

Substituting these values into the formula, we get:

A = π * 20.0 cm * 15.0 cm

Now, let's convert the units to meters:

A = π * 0.2 m * 0.15 m

Next, let's calculate the value of the area:

A = 0.09425 m²

Now, we can calculate the torque using the given values:

N = 8 turns
B = 2.07 * 10^-4 T
θ = 90° (since the magnetic field is directed toward the left of the screen, and the coil lies in the plane of the screen)

Substituting these values into the torque formula:

Torque = 8 * 0.09425 m² * 2.07 * 10^-4 T * sin(90°)

Since sin(90°) = 1, we can simplify the expression:

Torque = 8 * 0.09425 m² * 2.07 * 10^-4 T

Now, let's calculate the torque:

Torque = 0.001562 T·m²

Therefore, the magnitude of the torque on the coil is 0.001562 T·m².

See http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magmom.html

Torque = 6*Area*B*sin 90 = 6 pi*a*b*B

The coil area gets multiplied by 6 because of the number of loops

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