Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 2182

Unit of \(\mu_{0} \mathrm{c}\) is same as that of (A) current (B) resistance (C) electric charge (D) velocity

Short Answer

Expert verified
None of the given options (A, B, C, or D) has the same unit as the expression \(\mu_{0} \mathrm{c}\), which has the unit \(\frac{\text{T} \cdot \text{m}^{2}}{\text{A} \cdot \text{s}}\).

Step by step solution

01

Identify the components of the expression

In the expression \(\mu_{0} \mathrm{c}\), \(\mu_{0}\) is the permeability of free space value and \(\mathrm{c}\) is the speed of light.
02

Determine the units of each component

The permeability of free space, \(\mu_{0}\), is a constant and has a unit of \(\frac{\text{T} \cdot \text{m}}{\text{A}}\), where T is the unit for magnetic field strength (Tesla), m is the unit of length (meter), and A is the unit of electric current (Ampere). The speed of light, \(\mathrm{c}\), is also a constant and has a unit of \(\frac{\text{m}}{\text{s}}\), where m is the unit of length (meter) and s is the unit of time (second).
03

Determine the unit of the expression

Now, to find the unit of the expression \(\mu_{0} \mathrm{c}\), we simply need to multiply the units of each component: \[ \frac{\text{T} \cdot \text{m}}{\text{A}} \times \frac{\text{m}}{\text{s}} = \frac{\text{T} \cdot \text{m}^{2}}{\text{A} \cdot \text{s}} \]
04

Compare the unit of the expression with the given options

Finally, looking at the options, we check which one has the same unit as \(\frac{\text{T} \cdot \text{m}^{2}}{\text{A} \cdot \text{s}}\): (A) Current: unit is Ampere (A) (B) Resistance: unit is Ohm \(\left( \Omega = \frac{\text{V}}{\text{A}} \right)\) (C) Electric charge: unit is Coulomb (C) (D) Velocity: unit is \(\frac{\text{m}}{\text{s}}\) None of the given options has the same unit as the expression \(\mu_{0} \mathrm{c}\), which means that none of the options (A, B, C, or D) is correct.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The electric and magnetic field of an electromagnetic wave are (A) in phase and perpendicular to each other (B) in phase and parallel to each other (C) in opposite phase and perpendicular to each other (D) in opposite phase and parallel to each other

For an electromagnetic wave, the phase difference between vectors \(\mathrm{E}^{-}\) and \(\mathrm{B}^{-}\) (far away from the source) (A) 0 (B) \([\pi / 2]\) (C) \(\pi\) (D) \([3 \pi / 2]\)

An electromagnetic wave going through vacuum is described by $E=E_{0} \sin (k x-\cot )$. Which of the following is independent of the wavelength? (A) \(\omega\) (B) \((\mathrm{k} / \mathrm{c})\) (C) \(\mathrm{k}_{\mathfrak{e}}\) (D) \(\mathrm{k}\)

If the direction of magnetic field \(\mathrm{B}^{\rightarrow}\) at some instant is along + ve \(Z\) direction and the electromagnetic wave is propagating along + ve \(\mathrm{X}\) direction, then the direction of electric field \(\mathrm{E}^{\rightarrow}\) at that instant is (A) along - ve Y direction (B) along + ve Y direction (C) along + ve \(\mathrm{X}\) direction (B) along - ve \(\mathrm{X}\) direction

If the wavelength of light is \(4000^{\circ} \mathrm{A}\) then the number of waves in \(1 \mathrm{~mm}\) length will be (A) \(2.5\) (B) 2500 (C) 250 (D) 25000
See all solutions

Recommended explanations on English Textbooks

Lexis and Semantics

Read Explanation

Email

Read Explanation

Text Comparison

Read Explanation

Single Paragraph Essay

Read Explanation

Lexis and Semantics Summary

Read Explanation

Language Acquisition

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.

Sign-up for free