The relation between $\mu$ and $H$ for a specimen of iron is $\mu = [\frac{0.4}{H} + 12 \times 10^{-4}] \ H m^{-1}$. The value of $H$ which produces a flux density of $1 \ T$ will be ($\mu =$ magnetic permeability,$H =$ magnetic intensity).
- A$250 \ A m^{-1}$
- B$500 \ A m^{-1}$
- C$750 \ A m^{-1}$
- D$10^3 \ A m^{-1}$
Explore More
Similar Questions
If the relative permeability of iron is $5500$,then its magnetic susceptibility is:
$A$ rod of ferromagnetic material with dimensions $10 \, cm \times 0.5 \, cm \times 0.2 \, cm$ is placed in a magnetic field of strength $0.5 \times 10^4 \, A/m$. As a result,a magnetic moment of $5 \, A \cdot m^2$ is produced in the rod. The value of the total magnetic induction will be (in $Tesla$):
Difficult
View SolutionThe magnetic susceptibility of iron is $5499$. The relative permeability of iron will be
$A$ domain in a ferromagnetic substance is in the form of a cube of side $1 \mu m$. If it contains $8 \times 10^{10}$ atoms and each atomic dipole has a dipole moment of $9 \times 10^{-24} \ A \cdot m^{2}$,then the magnetisation of the domain is
$A$ toroid has an average diameter of $2.5\,m$,$400$ turns,and a current of $2\,A$. If the total magnetic field inside the toroid is $10\,T$,what is the intensity of magnetization (in $A/m$)?
MediumAIIMS 2019
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo