Examoogle
ExamsTest SeriesCBATRank CheckPrevious Year PapersPassBook StoreMy BooksAI Tutor
🛒0
अA
Examoogle

India's most trusted platform for competitive exam PDF books. Expert-authored, watermark-protected, instant access.

Exams & Practice
All Exams & SyllabusMock Test SeriesPrevious Year PapersPractice Questions (MCQs)Recruitment Notifications
Quick Links
Examoogle AI TutorExam NewsBook StoreMy BooksLogin / Sign Up
Support
About UsRefund PolicyPrivacy PolicyTerms of UseContact Us
© 2026 Examoogle. India's #1 competitive exam AI tutor.
🔒 SSL Secured📱 UPI Accepted🧾 GST Invoice
Examoogle

Join 60,000+ competitive exam aspirants

or with email
By continuing, you agree to ourTerms of Service&Privacy Policy
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Back to Practice Questions
ElectricalElectromagnetics Field Theory
PrevNext

For a given ferrite material operating in linear mode, B = 0.05 T and μr = 50. The magnetic field H is

A

mA/m

B

125 A/m

C

796 A/m

D

1000 A/m

Correct Answer

⚙️ TE • Technical Direct FormulaElectricalElectromagnetics Field Theory
Option C

796 A/m

Quick Summary:

Given: Magnetic flux density B = 0.05 T, relative permeability μr = 50, and permeability of free space μ0 = 4π × 10°-7 H/m.

📐MAMath SolutionDirect Formula
📋 Given

Magnetic flux density B = 0.05 T, relative permeability μr = 50, and permeability of free space μ0 = 4π × 10°-7 H/m.

🔢 Formula Used

H=Bμ0×μrH = \frac{B}{\mu_0 \times \mu_r}H=μ0​×μr​B​

🔢 Step-by-Step Solution
1

Identify the relationship between B and H

The magnetic field intensity HHH is related to the magnetic flux density BBB by the permeability of the material μ\muμ, where μ=μ0μr\mu = \mu_0 \mu_rμ=μ0​μr​.

B=μH=μ0μrHB = \mu H = \mu_0 \mu_r HB=μH=μ0​μr​H

2

Rearrange the formula to solve for H

Isolate HHH by dividing both sides by the permeability of the medium μ0μr\mu_0 \mu_rμ0​μr​.

H=Bμ0μrH = \frac{B}{\mu_0 \mu_r}H=μ0​μr​B​

3

Substitute the given values

Plug in B=0.05 TB = 0.05 \text{ T}B=0.05 T, μ0=4π×10°−7 H/m\mu_0 = 4\pi \times 10°{-7} \text{ H/m}μ0​=4π×10°−7 H/m, and μr=50\mu_r = 50μr​=50.

H=0.054π×10°−7×50H = \frac{0.05}{4\pi \times 10°{-7} \times 50}H=4π×10°−7×500.05​

4

Calculate the result

Simplify the expression: H=0.05200π×10°−7=0.050.00006283≈795.77 A/mH = \frac{0.05}{200\pi \times 10°{-7}} = \frac{0.05}{0.00006283} \approx 795.77 \text{ A/m}H=200π×10°−70.05​=0.000062830.05​≈795.77 A/m. Rounding to the nearest integer gives 796 A/m796 \text{ A/m}796 A/m.

H≈796 A/mH \approx 796 \text{ A/m}H≈796 A/m

✅

C is correct because calculating the magnetic field intensity using the formula H=Bμ0μrH = \frac{B}{\mu_0 \mu_r}H=μ0​μr​B​ yields approximately 796 A/m.

Core Concepts Used
Click any tag to open in AI Tutor
Magnetic field intensity Magnetic flux density Permeability of free space
💡 EXAM TIP

This formula is fundamental in transformer and motor design; remember that μ0\mu_0μ0​ is constant while μr\mu_rμr​ depends on the core material saturation characteristics.

Related Questions

ElectricalElectromagnetics Field Theory
For air, the Maxwell's equation hold true is
ElectricalElectromagnetics Field Theory
For any metals, the Maxwell's equation hold true is
ElectricalElectromagnetics Field Theory
Maxwell’s equation not be represented in
ElectricalElectromagnetics Field Theory
Maxwell's equation derived from Ampere’s law is
ElectricalElectromagnetics Field Theory
Maxwell's equation derived from Faraday’s law is

Discussion (0)

Loading discussion...
PrevNext