/*! 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 21 A \(50 \mathrm{mH}\) coil carrie... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 21

A \(50 \mathrm{mH}\) coil carries a current of \(2 \mathrm{~A}\), the energy stored in it in \(\mathrm{J}\) is (A) \(0.05\) (B) \(0.1\) (C) \(0.5\) (D) 1

Short Answer

Expert verified
The energy stored in a $50 \mathrm{mH}$ coil carrying a current of $2 \mathrm{A}$ is found using the formula \(W = \frac{1}{2}LI^2\), where \(W\) is the energy stored in the inductor, \(L\) is the inductance, and \(I\) is the current passing through the coil. Substituting the given values, we get \(W = \frac{1}{2} \times (50 \times 10^{-3}) \times (2)^{2} = 0.1 \mathrm{J}\). Therefore, the correct answer is (B) $0.1 \mathrm{J}$.

Step by step solution

01

Note down the given values

We have been given the following values: Inductance, \(L = 50~mH = 50 \times 10^{-3}~H\) Current, \(I = 2 ~A\)
02

Apply the formula to find the energy stored in the coil

Now, let's substitute the given values into the formula for energy stored in the inductor, which is \(W = \frac{1}{2}LI^2\). \(W = \frac{1}{2} \times (50 \times 10^{-3}) \times (2)^{2}\)
03

Calculate the energy in Joules

After substituting the values, we can solve the formula to find the energy in Joules: \(W = \frac{1}{2} \times 50 \times 10^{-3} \times 4\) \(W = \frac{200}{2} \times 10^{-3}\) \(W = 100 \times 10^{-3}\) \(W = 0.1~J\) So, the energy stored in the coil is \(0.1~J\). This corresponds to option (B).

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 91影视!

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 is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil, it starts oscillating; it is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to (A) development of air current when the plate is placed. (B) induction of electrical charge on the plate. (C) shielding of magnetic lines of force as aluminium is a paramagnetic material. (D) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.

In a coil when current changes from \(10 \mathrm{~A}\) to \(2 \mathrm{~A}\) in time \(0.1 \mathrm{~s}\), induced EMF is \(3.20 \mathrm{~V}\). The self-inductance of coil is (A) \(4 \mathrm{H}\) (B) \(0.4 \mathrm{H}\) (C) \(0.04 \mathrm{H}\) (D) \(5 \mathrm{H}\)

Figure \(16.66\) shows four rods having \(\lambda=0.5 \Omega / \mathrm{m}\) resistance per unit length. The arrangement is kept in a magnetic field of constant magnitude \(B=2 \mathrm{~T}\) and directed perpendicular to the plane of the figure and directed inwards. Initially, the rods form a square of side length \(\ell=15 \mathrm{~m}\) as shown. Now each wire starts moving with constant velocity \(v=5 \mathrm{~m} / \mathrm{s}\) towards opposite wire. Find the force required in newton on each wire to keep its velocity constant at \(t=1 \mathrm{~s}\).

Loop \(A\) of radius \(r(r \ll R)\) moves towards a constant current carrying loop \(B\) with a constant velocity \(v\) in such a way that their planes are parallel and coaxial. The distance between the loops when the induced EMF in loop \(A\) is maximum is (A) \(R\) (B) \(\frac{R}{\sqrt{2}}\) (C) \(\frac{R}{2}\) (D) \(R\left(1-\frac{1}{\sqrt{2}}\right)\)

The power factor of an AC circuit having resistance \((R)\) and inductance \((L)\) connected in series and an angular velocity \(\omega\) is (A) \(R / \omega L\) (B) \(R /\left(R^{2}+\omega^{2} L^{2}\right)^{1 / 2}\) (C) \(\omega L / R\) (D) \(R /\left(R^{2}-\omega^{2} L^{2}\right)^{1 / 2}\)
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Modern Physics

Read Explanation

Kinematics Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Electromagnetism

Read Explanation

Geometrical and Physical Optics

Read Explanation
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.