Question

A 1.80 cm × 1.80 cm square loop of wire with resistance 1.20×10^−2 Ω has one edge parallel to a long straight wire. The near edge of the loop is 1.20 cm from the wire. The current in the wire is increasing at the rate of 100 A/s . What is the current in the loop?

Answer 1

To find the current in the loop, we can use Faraday's law of electromagnetic induction. When the current in the straight wire changes, it induces an electromotive force (emf) in the loop. This emf creates a current in the loop, which is given by Ohm's law: I = V/R, where I is the current, V is the emf, and R is the resistance.

Step-by-step explanation:

To find the current in the loop, we can use Faraday's law of electromagnetic induction. When the current in the straight wire changes, it induces an electromotive force (emf) in the loop. This emf creates a current in the loop, which is given by Ohm's law: I = V/R, where I is the current, V is the emf, and R is the resistance. In this case, the emf induced in the loop can be found using Ampère's law: ∮B·dl = μ₀I_enclosed, where B is the magnetic field, dl is a differential distance along the loop, and I_enclosed is the enclosed current.

Using Ampère's law and assuming a clockwise current flow in the loop, we can find the magnetic field at the near edge of the loop by integrating B·dl. The integral becomes B(l_0 + l) - B(l_0), where l_0 is the distance from the near edge of the loop to the wire and l is the length of the loop. Since the magnetic field is constant along the loop and parallel to the edge, the integral simplifies to B(l_0 + l) - B(l_0) = B*l. Substituting this value into Ampère's law, we get B*l = μ₀I_enclosed. Solving for I_enclosed, we find I_enclosed = B*l/μ₀.

Now we can substitute the given values into the equation to find the current in the loop. The magnetic field is not given, so we cannot find the current without this information.

Answer 2

current in loop is 27.5 μA

Step-by-step explanation:

given data

side a = b = 1.80 cm = 0.018 m

resistance 1.20×10^−2 Ω

edge distance = 1.20 cm = 0.012 m

to find out

current in loop

solution

we know here rate dI/dt = 100 A/s

so here current = 1 / R × d∅/dt ..............a

and

magnetic flux ∅ = μ Ib / 2π × ln((a+c)/c) ..............b

differentiate it

d∅ /dt = μ b / 2π × ln((a+c)/c) dI/dt

now put all these value and find d∅ /dt

d∅ /dt = 4π
`10^(-7)` (0.018) / 2π × ln((0.012+0.018)/0.012) × 100

d∅ /dt = 3.3 ×
`10^(-7)` V

so

current = (d∅ /dt) / R

current = 3.3 ×
`10^(-7)` / 0.0120

current = 27.5 ×
`10^(-6)` A

so current is 27.5 μA