A long solenoid has 1400 turns per meter of length, and it carries a current of 2.0 A. A small circular coil of wire is placed inside the solenoid with the normal to the coil oriented at an angle of 90.0˚ with respect to the axis of the solenoid. The coil consists of 46 turns, has an area of 1.2 × 10-3 m2, and carries a current of 0.75 A. Find the torque exerted on the coil.
To find the torque exerted on the coil, we need to use the formula:
Torque (τ) = N * B * A * sin(θ)
where:
τ is the torque,
N is the number of turns in the coil,
B is the magnetic field strength,
A is the area of the coil, and
θ is the angle between the normal to the coil and the magnetic field.
First, let's find the magnetic field strength (B) inside the solenoid:
B = μ₀ * n * I
where:
B is the magnetic field strength,
μ₀ is the permeability of free space,
n is the number of turns per unit length in the solenoid, and
I is the current flowing through the solenoid.
Given that n = 1400 turns/m and I = 2.0 A, we can substitute these values in the formula to find B.
B = (4π × 10^-7 T·m/A)(1400 turns/m)(2.0 A)
B ≈ 1.76 × 10^-3 T
Next, we can calculate the torque exerted on the coil using the given values:
N = 46 turns
A = 1.2 × 10^-3 m²
θ = 90° (sin(90°) = 1)
τ = 46 * (1.76 × 10^-3 T) * (1.2 × 10^-3 m²) * 1
τ ≈ 1.003 × 10^-5 N·m
Therefore, the torque exerted on the coil is approximately 1.003 × 10^-5 N·m.