A 2.10 cm × 2.10 cm square loop of wire with resistance 1.20×10?2 ? has one edge

Question

A 2.10 cm × 2.10 cm square loop of wire with resistance 1.20×10?2 ? 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 120 A/s . What is the current in the loop in ?A?

« previous 2 of 6 | next >» Prayer 2 Part A A 2.10 cm × 2.10 cm square loop of wire with resistance 1.20×10-2 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 120 A/s What is the current in the loop? HA Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

As the question mentions that the current in the wire is increasing at a rate of 120 A/s. This would imply that the net magnetic flux passing through the square loop kept nearby would also be changing which in turn would lead to induction of an emf in the loop.

We know that for a closed loop, with changing magnetic flux, the emf induced is given as:

EMF = d / dt

Now as to determine the current in the loop, we will determine the relation of the flux through the loop and the current in the wire and then use the rate of the change of the flux to find the emf induced.

We know that the magnetic field due to an infinite wire at a distance r is given as:

B = o I / 2r

Let us assume a small section of the loop of thickness dr at a distance r from the wire.

Net flux through this section of width dx would be: d = o I (2.1) x10^-2 dr / 2r

We can integrate the above expression to determien the net flux through the loop as

= (o I / 2) [ln(3.3 / 1.2)] = I x 2.0232 x 10^-7 x 2.1 x 10^-2 Tm^2

Now the rate of change of flux = d/dt = 4.24872 x 10^-9 x dI/dt = 4.24872 x 10^-9 x 120 = 509.8464 x 10^-9

Therefore the emf induced in the loop would be: 509.8464 x 10^-9 V

Hence the current in the loop = EMF / Resistance = 509.8464 x 10^-9 / 1.2 x 10^-2 = 424.872 x 10^-7 amperes = 42.487 A

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