/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 In a region of space, a magnetic... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a region of space, a magnetic field points in the +\(x\)-direction (toward the right). Its magnitude varies with position according to the formula \(B_x = B_0 + bx\), where \(B_0\) and \(b\) are positive constants, for \(x \geq\) 0. A flat coil of area \(A\) moves with uniform speed \(v\) from right to left with the plane of its area always perpendicular to this field. (a) What is the emf induced in this coil while it is to the right of the origin? (b) As viewed from the origin, what is the direction (clockwise or counterclockwise) of the current induced in the coil? (c) If instead the coil moved from left to right, what would be the answers to parts (a) and (b)?

Short Answer

Expert verified
(a) \( Abv \), clockwise for right-left. If left-right, (a) \( Abv \), counterclockwise.

Step by step solution

01

Identify necessary formula

The formula for electromotive force (emf) induced in a coil moving in a magnetic field is given by Faraday's law of electromagnetic induction: \( \mathcal{E} = -\frac{d\Phi_B}{dt} \), where \( \Phi_B \) is the magnetic flux.
02

Calculate magnetic flux through the coil

The magnetic flux \( \Phi_B \) through the coil is calculated as \( \Phi_B = B_x \cdot A \). Here, \( B_x = B_0 + bx \) is the magnetic field and \( A \) is the coil's area. Thus, \( \Phi_B = (B_0 + bx)A \).
03

Determine the rate of change of flux

Since flux \( \Phi_B \) is \( (B_0 + bx)A \), the change in flux with respect to time as the coil moves is: \[ \frac{d\Phi_B}{dt} = A\frac{d(B_0 + bx)}{dt} = Ab\frac{dx}{dt} = Abv \], where \( \frac{dx}{dt} \) is the speed \( v \) of the coil.
04

Calculate the induced emf

Using Faraday's law, the induced emf is \( \mathcal{E} = - Abv \). The negative sign indicates the direction according to Lenz's law, but the magnitude of the emf is \( Abv \).
05

Determine the current direction for right-moving coil

When viewed from the origin and the coil is moving from right to left, the induced current, due to Lenz's law, will be in the clockwise direction to oppose the decrease in flux as it moves away from the origin.
06

Reevaluating for coil moving left-to-right

If the coil moves from left to right, the expressions derived for emf magnitude remain the same: \( \mathcal{E} = Abv \).
07

Determine the current direction for left-moving coil

Moving from left to right towards the origin, the induced current will be counterclockwise as viewed from the origin. This is because as the coil enters regions of stronger magnetic field, it opposes the increase in magnetic flux.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Faraday's Law
Faraday's Law is a fundamental principle of electromagnetism which states that a change in magnetic flux through a coil will induce an electromotive force (emf) in the wire. This phenomenon is called electromagnetic induction. The law is mathematically represented by the formula:
  • \( \mathcal{E} = -\frac{d\Phi_B}{dt} \)
where \( \mathcal{E} \) is the induced emf and \( \Phi_B \) is the magnetic flux. Faraday's Law highlights the relationship between the rate of change of magnetic flux and the resulting emf.
This change can be due to alterations in the magnetic field strength, the area of the coil, or its orientation relative to the magnetic field. Negative sign in Faraday's Law comes from Lenz's law, explaining the direction of the induced current.
In practical terms, Faraday's Law forms the basis for electric generators, transformers, and many other technologies that involve changing magnetic fields to generate electricity.
Magnetic Flux
Magnetic flux, denoted as \( \Phi_B \), is a measure of the number of magnetic field lines that pass through a given area. It's an important concept when dealing with electromagnetic induction as it determines the magnitude of induced emf according to Faraday's Law. The magnetic flux through a surface is calculated as:
  • \( \Phi_B = B \cdot A \cdot \cos(\theta) \)
where \( B \) is the magnetic field strength, \( A \) is the area through which the field lines pass, and \( \theta \) is the angle between the field lines and the perpendicular to the surface.
When a coil moves in a non-uniform magnetic field, changes in magnetic flux can be the result of:
  • changes in \( B \)
  • the coil's position \( x \)
  • the orientation of the coil relative to the field.
In the given exercise, the magnetic flux was calculated using the expression \( \Phi_B = (B_0 + bx)A \) showing how flux varies with the coil's position in a non-uniform field. This impacts the induced current based on how the coil moves relative to the changing field strength.
Lenz's Law
Lenz's Law describes the directionality of induced currents. It's essentially a manifestation of the conservation of energy principle, ensuring that the induced current will always oppose the motion or change causing it. In mathematical terms, this is represented as the negative sign in Faraday's Law:
  • \( \mathcal{E} = -\frac{d\Phi_B}{dt} \)
Lenz's Law states that the direction of any induced current will be such that the magnetic field it creates will oppose the change in the original magnetic flux. This means:
  • If a magnetic field through a coil increases, the induced current produces its own magnetic field opposing this increase.
  • Conversely, if the magnetic field through a coil decreases, the induced current generates a magnetic field trying to maintain it.
In the textbook example, when the coil moves from right to left (as viewed from the origin), the magnetic flux through the coil decreases as the coil moves away, and Lenz’s Law predicts a clockwise induced current, trying to counteract this decrease. Conversely, moving from left to right causes an increase in flux, leading to a counterclockwise current.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A circular conducting ring with radius \(r_0 =\) 0.0420 m lies in the xy-plane in a region of uniform magnetic field \(\overrightarrow{B} = B_0 [1 - 3(t/t_0)^2 + 2(t/t_0)^3]\hat{k}\). In this expression, \(t_0 =\) 0.0100 s and is constant, \(t\) is time, \(\hat{k}\) is the unit vector in the +\(z\)-direction, and \(B_0\) = 0.0800 T and is constant. At points \(a\) and \(b\) (Fig. P29.58) there is a small gap in the ring with wires leading to an external circuit of resistance \(R =\) 12.0 \(\Omega\). There is no magnetic field at the location of the external circuit. (a) Derive an expression, as a function of time, for the total magnetic flux \(\Phi_B\) through the ring. (b) Determine the emf induced in the ring at time \(t =\) 5.00 \(\times\) 10\(^{-3}\) s. What is the polarity of the emf? (c) Because of the internal resistance of the ring, the current through \(R\) at the time given in part (b) is only 3.00 mA. Determine the internal resistance of the ring. (d) Determine the emf in the ring at a time \(t =\) 1.21 \(\times\) 10\(^{-2}\) s. What is the polarity of the emf? (e) Determine the time at which the current through \(R\) reverses its direction.

A single loop of wire with an area of 0.0900 m\(^2\) is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.190 T/s. (a) What emf is induced in this loop? (b) If the loop has a resistance of 0.600 \(\Omega\), find the current induced in the loop.

At temperatures near absolute zero, \(B_c\) approaches 0.142 T for vanadium, a type-I superconductor. The normal phase of vanadium has a magnetic susceptibility close to zero. Consider a long, thin vanadium cylinder with its axis parallel to an external magnetic field \(\overrightarrow{B}_0\) in the +\(x\)-direction. At points far from the ends of the cylinder, by symmetry, all the magnetic vectors are parallel to the x-axis. At temperatures near absolute zero, what are the resultant magnetic field \(\overrightarrow{B}\) and the magnetization \(\overrightarrow{M}\) inside and outside the cylinder (far from the ends) for (a) \(\overrightarrow{B}_0\) = (0.130 T)\(\hat{\imath}\) and (b) \(\overrightarrow{B}_0\) = (0.260 T)\(\hat{\imath}\) ?

A slender rod, 0.240 m long, rotates with an angular speed of 8.80 rad/s about an axis through one end and perpendicular to the rod. The plane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of 0.650 T. (a) What is the induced emf in the rod? (b) What is the potential difference between its ends? (c) Suppose instead the rod rotates at 8.80 rad/s about an axis through its center and perpendicular to the rod. In this case, what is the potential difference between the ends of the rod? Between the center of the rod and one end?

A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a constant rate of 12.0 cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop and with magnitude 0.500 T. (a) Find the emf induced in the loop at the instant when 9.0 s have passed. (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.