Question

A square loop whose sides are 2 cm long is made with copper wire of radius 8 mm, assuming resistivity of copper is 1.72 x 10-8 Ohm X m. If a magnetic field perpendicular to the loop is changing at a constant rate of 3 mT/s, what is the current in the loop?

Answer 1

To determine the current induced in a square copper loop by a changing magnetic field, apply Faraday's Law to calculate induced emf, and Ohm's Law relating current to emf and resistance, taking into account the resistivity and dimensions of the copper wire.

Step-by-step explanation:

To find the current induced in a copper wire square loop due to a changing magnetic field, we use Faraday's Law of electromagnetic induction which states that 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. First, we calculate the induced emf using the formula emf = -dΦ/dt where Φ is the magnetic flux and dt is the change in time. The magnetic flux Φ is given by the product of the magnetic field B and the area A of the loop (assuming the field is perpendicular to the loop), so Φ = B × A. With a changing magnetic field, the rate of change of Φ is then dB/dt times the area A. Once we have the induced emf, we can find the current I by using Ohm's Law, which is I = emf / R, where R is the resistance of the loop.

The resistance R of the loop can be calculated using the resistivity ρ of copper, the length l of the wire, and the cross-sectional area A of the wire using the formula R = ρ × (l/A). The length of the wire is the perimeter of the square loop, and the cross-sectional area can be calculated from the radius of the copper wire. Finally, after calculating R and the induced emf, we can divide the emf by R to find the induced current I.

Answer 2

The current in the loop is 1.2 µA.

Step-by-step explanation:

To determine the current in the loop, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the induced current in a loop is equal to the rate of change of magnetic flux through the loop. The magnetic flux is given by the product of the magnetic field and the area of the loop.

In this case, the magnetic field is changing at a rate of 3 mT/s, which is equivalent to 3 x 10⁻³ T/s. The area of the loop is equal to the square of the side length, so it is (2 cm)² = 4 cm². Converting to square meters, we get 4 cm² = 4 x 10⁻⁴ m².

Therefore, the rate of change of magnetic flux is (3 x 10⁻³ T/s)(4 x 10⁻⁴ m²) = 1.2 x 10⁻⁶ Wb/s. Since the resistivity of copper is given as 1.72 x 10⁸ Ohm X m, we can use Ohm's law to calculate the current: I = V/R. The induced electromotive force (emf) V is equal to the rate of change of magnetic flux, so V = (1.2 x 10⁻⁶ Wb/s)(1.72 x 10⁻⁸ Ohm X m). The resistance R is given as 1.72 x 10⁻⁸ Ohm X m.

Substituting the values, we get I = (1.2 x 10⁻⁶ Wb/s)(1.72 x 10⁻⁸ Ohm X m) / (1.72 x 10⁻⁸ Ohm X m) = 1.2 x 10⁻⁶ A = 1.2 µA.