/*! 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 1 The magnetic flux \(\phi\) (in w... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The magnetic flux \(\phi\) (in weber) in a closed circuit of resistance 10 ohm varies with time \(t\) (in second) according to equation \(\phi=6 t^{2}-5 t+1\). The magnitude of induced current at \(t=0.25 \mathrm{~s}\) is: (a) \(1.2 \mathrm{~A}\) (b) \(0.8 \mathrm{~A}\) (c) \(0.6 \mathrm{~A}\) (d) \(0.2 \mathrm{~A}\)

Short Answer

Expert verified
The induced current is 0.2 A.

Step by step solution

01

Understand the Problem

We need to find the magnitude of the induced current in a circuit given the expression for magnetic flux. The flux varies with time as \( \phi = 6t^2 - 5t + 1 \). The circuit has a resistance of 10 ohms, and we need to determine the induced current at \( t = 0.25 \) seconds.
02

Apply Faraday's Law of Induction

Faraday's Law states that the induced electromotive force (emf) in a circuit is equal to the negative rate of change of magnetic flux through the circuit. The formula is \( \varepsilon = -\frac{d\phi}{dt} \). We need to compute \( \frac{d\phi}{dt} \) and evaluate it at \( t = 0.25 \) seconds.
03

Differentiate Magnetic Flux

Differentiate \( \phi = 6t^2 - 5t + 1 \) with respect to \( t \).\[ \frac{d\phi}{dt} = \frac{d}{dt}(6t^2) - \frac{d}{dt}(5t) + \frac{d}{dt}(1) \]\[ \frac{d\phi}{dt} = 12t - 5 \]
04

Evaluate the Derivative at t = 0.25

Substitute \( t = 0.25 \) into the derivative:\[ \frac{d\phi}{dt} = 12(0.25) - 5 = 3 - 5 = -2 \]Thus, the induced emf is \( \varepsilon = -(-2) = 2 \) volts.
05

Calculate Induced Current

Use Ohm's Law, which states \( I = \frac{\varepsilon}{R} \), where \( I \) is the induced current and \( R \) is the resistance:\[ I = \frac{2}{10} = 0.2 \text{ A} \]
06

Conclusion

The magnitude of the induced current at \( t = 0.25 \) seconds is \( 0.2 \) A.

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.

Magnetic Flux
Magnetic flux is a measure of the magnetic field passing through a given area. Imagine it as the number of magnetic lines crossing through a surface. It is represented by the symbol \( \phi \) and is measured in units of Weber (Wb). Magnetic flux depends on two key factors:
  • The strength of the magnetic field (B)
  • The area (A) of the surface perpendicular to the field
Together, it’s expressed mathematically as \( \phi = B \times A \times \cos(\theta) \), where \( \theta \) is the angle between the magnetic field and the normal (perpendicular) to the surface.In the context of the given exercise, magnetic flux varies with time according to the equation \( \phi = 6t^2 - 5t + 1 \). This equation implies that the flux changes as the time \( t \) changes. The goal in the problem is to find out how this changing flux induces an electromotive force (emf) which results in a current flowing through a circuit.
Ohm's Law
Ohm's Law is a fundamental principle in electrical circuits, relating the current flow through a conductor to the voltage across it and its resistance. It is expressed by the formula:\[ I = \frac{V}{R} \]where:
  • \( I \) is the current in amperes (A)
  • \( V \) is the voltage in volts (V)
  • \( R \) is the resistance in ohms (\( \Omega \))
In this exercise, following Faraday's Law of Induction, once we have found the induced emf in the circuit as 2 volts, Ohm’s Law helps us determine the current. With a given resistance of 10 ohms, the induced current is found by rearranging the formula: \( I = \frac{\varepsilon}{R} = \frac{2}{10} = 0.2 \) amperes. This means the induced current at \( t = 0.25 \) seconds is 0.2 amperes. Ohm’s Law is invaluable in circuit calculations, providing a straightforward way to relate these three key quantities.
Induced Electromotive Force
The induced electromotive force, often abbreviated as emf, is a result of changing magnetic fields within a closed loop of wire, according to Faraday's Law of Induction. It is the driving force that causes electrons to move and create an electric current. Mathematically, Faraday's Law is expressed as:\[ \varepsilon = -\frac{d\phi}{dt} \]Here, \( \varepsilon \) represents the induced emf, and \( \frac{d\phi}{dt} \) is the rate of change of magnetic flux. The negative sign is a reflection of Lenz's Law, indicating that the direction of induced emf and current will oppose the change in flux. This negative sign is crucial because it ensures the conservation of energy in electromagnetic processes.In the problem, after differentiating the flux equation \( \phi = 6t^2 - 5t + 1 \), the change in flux \( \frac{d\phi}{dt} \) at \( t = 0.25 \) seconds is calculated as \(-2\). Thus, the induced emf \( \varepsilon = 2 \) volts. The positive value for the emf tells us about the magnitude of the force inducing current around the circuit. Understanding how an emf is induced is essential for comprehending the workings of generators, transformers, and various electrical devices.

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 coil of inductance \(L=300 \mathrm{mH}\) and resistance \(R=140 \mathrm{~m} \Omega\) is connected to a constant voltage source. Current in the coil will reach to \(50 \%\) of its steady value after time \(t\) equals to: (a) \(155 \mathrm{~s}\) (b) \(0.755 \mathrm{~s}\) (c) \(0.155 \mathrm{~s}\) (d) \(1.48 \mathrm{~s}\)

A fan blade of length \(1 / \sqrt{\pi}\) meter rotates with frequency 5 cycle per second perpendicular to a magnetic field 10 tesla. What is potential difference between the centre and the end of blade ? (a) \(-50 \mathrm{~V}\) (b) \(+50 \mathrm{~V}\) (c) \(-2.0 \mathrm{~V}\) (d) \(+0.02 \mathrm{~V}\)

An Indian ship with a vertical conducting mass navigates the Indian ocean in the latitude of magnetic equator. To induce the greatest emf in the mast, the ship should proceed : (a) northward (b) southward (c) eastward (d) none of these

Two straight super-conducting rails form an angle \(\theta\) where their ends are joined a conducting bar having \(R_{0}\) resistance per unit length in contact with the rails and forming an isosceles triangle with them. The bar starts at the vertex at time \(t=0\) and moves with constant velocity \(v\) to right. A magnetic field \(B\) is present into the region (shown in figure). Find the force exerted by external agent to maintain constant velocity to the rod: (a) \(\frac{2 B^{2} v^{2} t}{R_{0}} \tan \frac{\theta}{2}\) (b) \(\frac{B^{2} v^{2} t}{R_{0}} \tan \frac{\theta}{2}\) (c) \(\frac{B^{2} v^{2} t}{R_{0}}\) (d) none of these

Two flat horizontal coils are mounted as shown. Which one of the following actions will not cause the sensitive galvanometer \(G\) to deflect? (a) Coils stationary and coil 1 moves upwards with \(K_{1}\) and \(K_{2}\) closed (b) Both coils stationary, \(K_{2}\) closed and \(K_{1}\) switched on or off (c) With \(K_{1}\) and \(K_{2}\) closed, a variable resistance \(R\) is increased and decreased rapidly (d) Both coils stationary, \(K_{1}\) closed and \(K_{2}\) switched on or off

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.