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A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0 ? with the direction of the field. when the magnetic field is increased uniformly from 200 ?t to 600 ?t in 0.400 s, an emf of magnitude of 80.0 mv is induced in the coil. what is the total length of the wire?

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To find the total length of wire, Faraday's Law is applied using the provided values for induced emf, magnetic field change, and the number of turns. The area of the square-shaped coil is computed, enabling the determination of the side length and consequently, the total length of wire by accounting for all four sides and the number of turns.

The student's question involves a magnetic field and electrical induction, which are related to Faraday's Law of electromagnetic induction. According to Faraday's Law, the induced electromotive force (emf) in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit. The formula to calculate the induced emf in the coil is given by ε = -N(dΦ/dt), where ε is the emf, N is the number of turns, and dΦ/dt represents the rate of change of magnetic flux. The flux change, Φ, is calculated by B*A*cos(θ), where B is the magnetic field strength, A is the area of the coil, and θ is the angle between the magnetic field vector and the normal to the coil's plane.

Since we are given the induced emf (ε = 80.0 mV), the number of turns (N = 50), the initial and final magnetic field (B - B = 600 μt - 200 μt), and the time interval (Δt = 0.400 s), we can use these to calculate the area, A, of the coil. Once the area is found, using the fact that the coil is square-shaped, we can calculate the side length and, from there, the total length of wire by multiplying the side length by 4 (since a square has four sides) and again by the number of turns (50).

To find A, we rearrange the emf equation to solve for A and then proceed with finding the side length and total length of wire. The total length will then be L = 4 * side length * number of turns.

User AllirionX
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The induced voltage is N(-dΦ/dt) where N=50 and flux is Φ=BAcos(θ). B is the field and A is the area. The trick is to see that because the loop is a square with side length L, that the area is L^2. For your magnetic flux density values I'm seeing "200 ?t" and 600 ?t" Am I right in thinking these are microTesla, uT? I will assume so but if it's something else you'll have to find the answer again So dB/dt = (600e-6-200e-6)/0.4=1e-3. Now,
80mV=(50)(1e-3)(L^2)cos(30)
L = 1.359m
Thus the perimeter of the square is 4(1.359m) = 5.437m
And the total length is (50)(5.437m) = 271.85m
User Harshal Kshatriya
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