An infinite wire carries an oscillating current I=I_0 \cos \omega t. A square loop wires with...

Question:

An infinite wire carries an oscillating current {eq}I=I_0 \cos \omega t {/eq}. A square loop wires with sides of length {eq}a {/eq} and resistance {eq}R {/eq} is located a distance {eq}a {/eq} from the infinite wire.

a. When the magnetic flux through the loop is at its maximum value, {eq}\phi_{max} {/eq}, what is the magnitude of the current in the loop?

b. Suppose at time {eq}t_0 {/eq} the current in the wire flows downward and is decreasing. What is the direction of forces between the infinite wire and the loop?

c. At the time {eq}t_0 {/eq}, the potential energy of the loop in the magnetic field created by the wire is

i. {eq}0 {/eq}

ii. Repulsive

iii. Attractive

Ampere's and Faraday's Laws

Ampere's law in its integral form states that the circulation of magnetic field around a closed loop is related to the amount of net current enclosed (flowing through) the loop and is given by:

{eq}\int \ \vec{B} \ \cdot \ \vec{dL} \ = \ \mu_0 \times I {/eq}

Where:

{eq}\vec{B} {/eq} - is the magnetic field in the region

{eq}\mu_0 \ = \ 1.257 \times 10^{-6} \ m \ kg \ s^{-2} \ A^{-2} {/eq} - is the permeability of free space.

Faraday's law of electromagnetic induction states that a varying magnetic flux through a closed loop induces an emf in the loop whose magnitude is given by:

{eq}E \ = \ \dfrac{d \phi}{dt} {/eq}

Where:

{eq}\phi \ = \ \int \ \vec{B} \cdot \ \vec{dA} {/eq} - is the magnetic flux through the loop.

Answer and Explanation:

Given:

  • Current through the infinite wire: {eq}I \ = \ I_0 \ \cos \omega t {/eq}
  • Side length of the square: {eq}a {/eq}
  • Distance from wire to the square side: {eq}a {/eq}
  • Resistance of the square loop: {eq}R {/eq}

Consider a small rectangular strip of length {eq}a {/eq} and width {eq}dx {/eq} at a distance {eq}x {/eq} from the wire. Magnetic field at this location can be computed using Ampere's law as:

{eq}\int \ B dL \ = \ \mu_0 \times I \\ \implies \ B \ = \ \dfrac{\mu_0 \times I}{2 \ \pi \times x} {/eq}

Magnetic flux through this elemental strip is: {eq}d \phi \ = \ B \ dA \ = \ B \times a \times dx \ = \ \dfrac{\mu_0 \times I \times a}{2 \pi x} \ dx {/eq}

Flux through the entire loop can be computed by integrating above equation from {eq}x \ = \ a {/eq} to {eq}x \ = 2 a {/eq}. Therefore,

{eq}\phi \ = \ \int^{2a}_a \ \dfrac{\mu_0 \times I \times a}{2 \pi \times x} \ dx = \ \dfrac{\mu_0 \times I \times a}{2 \pi} \times \ln 2 \ T m^2 {/eq}

a.

Magnetic flux through the loop is maximum when the current in the loop is maximum i.e. {eq}I \ = \ I_0 {/eq}

b.

When the current in the infinite wire is downwards, magnetic field due to this current in the loop will be out of the plane of the page.

From Lenz's law, induced current in the loop will be such that it opposes the change that caused it. When the current in the wire is decreasing, the magnetic flux through the loop is decreasing. Therefore, induced current will be such that it increases the magnetic flux through the loop i.e. induced current will be anti-clockwise and hence producing a magnetic field out of the plane of page.

As the force experienced by a wire in magnetic field is proportional to {eq}\vec{I} \ \times \ \vec{B} {/eq}, force on the loop will be to the right i.e. away from the wire.

c.

At {eq}t_0 {/eq}, as computed above, as the current induced in the loop is anti-clockwise and the magnetic field is out of the plane of the page, magnetic dipole moment of the loop and magnetic field are parallel to one another. As potential energy of the system is given by: {eq}U \ = \ - \mu \times B {/eq}, potential energy at this instant is negative.


Learn more about this topic:

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Faraday's Law of Electromagnetic Induction: Equation and Application

from High School Physics: Help and Review

Chapter 13 / Lesson 10
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