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Chapter 29: Problem 61

The inductance of a coil in which a current of \(0.1 \mathrm{~A}\) increasing at the rate of \(0.5 \mathrm{~A} / \mathrm{s}\) represents a power flow of \(\frac{1}{2}\) watt, is: (a) \(2 \mathrm{H}\) (b) \(8 \mathrm{H}\) (c) \(20 \mathrm{H}\) (d) \(10 \mathrm{H}\)

Short Answer

Expert verified
The correct inductance is 10 H, which matches option (d).

Step by step solution

01

Understand the Formula for Power

The power (P) associated with a changing current in an inductance (L) is given by the formula:\[ P = L \cdot \frac{di}{dt} \cdot i \] where \(\frac{di}{dt}\) is the rate of change of current and \(i\) is the current.
02

Substitute Given Values

From the problem, we have: \(P = \frac{1}{2} \text{ watt}\), \(i = 0.1 \text{ A}\), and \(\frac{di}{dt} = 0.5 \text{ A/s}\). Substitute these values into the formula:\[ \frac{1}{2} = L \cdot 0.5 \cdot 0.1 \]
03

Simplify the Equation

Simplify the expression on the right side of the equation to solve for \(L\):\[ \frac{1}{2} = L \cdot 0.05 \]
04

Solve for Inductance L

Divide both sides of the equation by \(0.05\) to isolate \(L\):\[ L = \frac{\frac{1}{2}}{0.05} \]\[ L = 10 \text{ H}\]
05

Match Solution to the Options

Compare the calculated inductance \(L = 10 \text{ H}\) with the options provided. The correct choice is (d) \(10 \text{ H}\).

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Key Concepts

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

Current Change Rate
When discussing inductors and electromagnetic fields, understanding the current change rate is crucial. The current change rate is represented by \( \frac{di}{dt} \), which denotes how quickly the current through an inductor changes over time. In our exercise, this rate is given as \(0.5 \text{ A/s}\). This means that for every second, the current increases by \(0.5\) amperes.

It's important to pay attention to the units here. Amperes (A) measure the amount of electric charge flowing per second, whereas the expression \(0.5 \text{ A/s} \) indicates a change in this current flow over each second.

Mastering this concept is important as it plays a direct role in calculating the energy interactions within inductors. The faster the current changes, the more impactful the electromagnetic effects become.
Power in Inductors
In an inductor, power is associated with how the electrical energy is stored and released. The power \( P \) can be calculated using the formula \( P = L \cdot \frac{di}{dt} \cdot i \), which links power to the inductance \(L\), the rate of current change \(\frac{di}{dt}\), and the current \(i\).

In the context of our problem, the power is specifically given as \( \frac{1}{2} \) watt. This indicates how intense the energy flow is within the inductor at the moment the current is changing.
  • The power reflects how quickly the energy is either being stored or delivered back to the circuit by the electromagnetic field.
  • This relationship showcases that both the current and how rapidly it changes impact the power within the inductor.
Being able to determine this is essential for understanding the behavior of circuits with inductance.
Inductance Calculation
Inductance, denoted by \(L\), is a measure of how effectively an inductor can store energy in a magnetic field when current flows through it. Our main goal is to calculate this value given specific problem conditions.

We begin by knowing:
  • Power \( P = \frac{1}{2} \text{ watt} \)
  • Current \( i = 0.1 \text{ A} \)
  • Rate of change \( \frac{di}{dt} = 0.5 \text{ A/s} \)
Using the formula \( P = L \cdot \frac{di}{dt} \cdot i \), we substitute these values to isolate \(L\):

\[\frac{1}{2} = L \cdot 0.5 \cdot 0.1\]
\[\frac{1}{2} = L \cdot 0.05\]
Divide both sides by \(0.05\) to solve for \(L\):
\[L = \frac{0.5}{0.05} = 10 \text{ H}\] This result demonstrates that the inductance is \(10 \text{ H}\), matching the correct answer from the options given.
Physics Problem Solving
Physics problem solving often requires breaking down a problem into smaller, manageable parts. In this exercise about inductors, the key steps provided a clear pathway to the solution.

Initially, understanding the relationship between power, current change rate, and current is crucial. By substituting known values, you can progressively simplify the equation.
  • Identify key given quantities and their units.
  • Apply the relevant formula, ensuring all units are consistent.
  • Simplify and solve step-by-step, verifying each step aligns with physical principles.
Problem solving, particularly in physics, often involves not just mathematical calculations but a deep comprehension of the physical phenomena. This approach ensures accurate results, especially in complex problems involving electromagnetism.

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Most popular questions from this chapter

The loop \(A B C D\) is moving with velocity ' \(v^{\prime}\) towards right. The magnetic field is \(4 \mathrm{~T}\). The loop is connected to a resistance of \(8 \mathrm{ohm}\). If steady current of \(2 \mathrm{~A}\) flows in the loop then value of \(v\) if loop has a resistance of 4 ohm, is : (Given \(A B=30 \mathrm{~cm}, A D=30 \mathrm{~cm}\) ) (a) \(\frac{50}{3} \mathrm{~m} / \mathrm{s}\) (b) \(20 \mathrm{~m} / \mathrm{s}\) (c) \(10 \mathrm{~m} / \mathrm{s}\) (d) \(\frac{100}{3} \mathrm{~m} / \mathrm{s}\)

In a cylindrical region, \(\vec{B}\) is static and uniform and points along the axis of the cylinder. Consider an equilateral triangle \(P Q R\) with its plane perpendicular to \(\vec{B}\). If \(B\) increases at a constant rate of \(1 \mathrm{~T} / \mathrm{s}\) and \(P Q=1 \mathrm{~m}\), the work done by induced electric force on a unit positive charge \((+1 C)\) taken from \(P\) to \(Q\) is : (a) positive (b) zero (c) \(1 / \sqrt{3} \mathrm{~J}\) (d) \(-1 / 2 \sqrt{3} \mathrm{~J}\)

A square loop lying in a \(\times\) * perpendicular magnetic field is \(\times \times\) changed in circle. If side of square is \(a\) and change occurs in \(t\) seconds \(\times\) in magnetic field \(B\) tesla, the induced emf is : (a) \(\frac{4}{\pi} \frac{B a^{2}}{t}\) (b) \(\frac{B a^{2}}{t}\) (c) \(\frac{B a^{2}}{t}\left(\frac{4}{\pi}-1\right)\) (d) zero

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

A circular loop of wire radius \(R\) rotates about \(z\) -axis with angular velocity \(\omega .\) The normal to the loop is perpendicular to z-axis. At \(t=0\) normal is parallel to \(y\) -axis. An external magnetic field \(\overrightarrow{\mathrm{B}}=B_{y} \hat{j}+B_{z} \hat{k}\) is applied. The emf induced in the coil will be: (a) \(\pi r^{2} \omega B_{y} \sin \omega t\) (b) \(\pi r^{2} \omega B_{z} \sin \omega t\) (c) \(\pi r^{2} \omega B_{z} \cos \omega t\) (d) \(\pi r^{2} \omega B_{y} \cos \omega t\)
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