
The relation between magnetic permeability , $\mu $ and field strength, $H$ for a specimen of iron is as follows $\mu $ = $\left( {\dfrac{{0.4}}{H} + 12 \times {{10}^{^{ - 4}}}} \right)$ $Henry/meter$ . The value of $H$ which produces flux density of $1T$ will be?
A) $250$ $A/m$
B) $500$ $A/m$
C) $750$ $A/m$
D) ${10^3}$ $A/m$
Answer
126.6k+ views
Hint: Given question has three terms magnetic permeability, field strength and flux density thus we will have to use a mathematical relation between these three which is, Flux density is a product of permeability and field strength.
Formula Used:
$B$ = $\mu$ $H$
Complete step by step answer:
In the question we have been a relation between Magnetic permeability and field strength but we need to find the value of field strength which will produce 1T flux density , Thus we will be using a direct relation between above three terms which is given by :
$B$ = $\mu$ $H$
Where B is flux density
Now we have given a relation is $\mu $= $\left( {\dfrac{{0.4}}{H} + 12 \times {{10}^{^{ - 4}}}} \right)$ with B = 1 ,
Therefore substituting the value of B and expression of $\mu $ in the main formula 1
$ \Rightarrow $1 = $\left( {\dfrac{{0.4}}{H} + 12 \times {{10}^{^{ - 4}}}} \right)$H
$ \Rightarrow $1 = $\left( {\dfrac{{0.4 + 12 \times {{10}^{ - 4}}H}}{H}} \right)H$
Cancelling H we get
$ \Rightarrow $$1 - 0.4 = 12 \times {10^{ - 4}}H$
$ \Rightarrow $$H = \dfrac{{.6}}{{12 \times {{10}^{ - 4}}}}$
$ \Rightarrow $$H = \dfrac{{1000 \times 6}}
{12} \\
\\
$
$ \Rightarrow H = 500$ $A/m$
Hence option B is the correct answer.
Note: Most important here is to remember the relation between the terms properly so that we don’t waste a lot of in straight formula based questions . It is also important that we have an idea of all the terms used in the problem so as to understand the process. It is important not to get confused within the terms themselves that is we should be very clear with denotion.
Formula Used:
$B$ = $\mu$ $H$
Complete step by step answer:
In the question we have been a relation between Magnetic permeability and field strength but we need to find the value of field strength which will produce 1T flux density , Thus we will be using a direct relation between above three terms which is given by :
$B$ = $\mu$ $H$
Where B is flux density
Now we have given a relation is $\mu $= $\left( {\dfrac{{0.4}}{H} + 12 \times {{10}^{^{ - 4}}}} \right)$ with B = 1 ,
Therefore substituting the value of B and expression of $\mu $ in the main formula 1
$ \Rightarrow $1 = $\left( {\dfrac{{0.4}}{H} + 12 \times {{10}^{^{ - 4}}}} \right)$H
$ \Rightarrow $1 = $\left( {\dfrac{{0.4 + 12 \times {{10}^{ - 4}}H}}{H}} \right)H$
Cancelling H we get
$ \Rightarrow $$1 - 0.4 = 12 \times {10^{ - 4}}H$
$ \Rightarrow $$H = \dfrac{{.6}}{{12 \times {{10}^{ - 4}}}}$
$ \Rightarrow $$H = \dfrac{{1000 \times 6}}
{12} \\
\\
$
$ \Rightarrow H = 500$ $A/m$
Hence option B is the correct answer.
Note: Most important here is to remember the relation between the terms properly so that we don’t waste a lot of in straight formula based questions . It is also important that we have an idea of all the terms used in the problem so as to understand the process. It is important not to get confused within the terms themselves that is we should be very clear with denotion.
Recently Updated Pages
Wheatstone Bridge - Working Principle, Formula, Derivation, Application

Young's Double Slit Experiment Step by Step Derivation

JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Login 2045: Step-by-Step Instructions and Details

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main
