Question

The mass of an iron rod is 80 gm and its magnetic movement is 10 Am 2 ${Am}^2$ when kept in external magnetizing field of strength If the density of iron is 8 gm/c.c. Then the value of intensity of magnetization will be (Take relative permeability of iron 10001)

• A. ${10}^6$ A /m
• B. ${10}^4$ A /m
• C. ${10}^2$ A /m
• D. 10 A/m

Answer 1

Answer: (A)

${10}^6$ A /m

$\text{Given}:$m=80gm, M=$10A{m}^2$, $\rho$=8gm/c.c.

$\text{Solution}$:

Intensity of magnetization, is given by

$I=\frac{MagneticMoment}{Volume}$= $\frac{M}{V}$

Also, Density $\rho$=$\frac{Mass\left(m\right)}{Volume\left(V\right)}$

$\therefore V=\frac{Mass}{Density}$= $\frac{80}{8}$=$10c{m}^3$= ${10}^{-5}{m}^3$

Hence,

I=$\frac{10}{{10}^{-5}}$= ${10}^6$ A/m

$\text{Hence A is the correct option}$