An iron bar having a cross-sectional area of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The magnetic flux $\phi$ is given by $\phi = B \cdot A$, where $B$ is the magnetic field and $A$ is the cross-sectional area.$B = \frac{\phi}{A} =

Vedclass · Accepted Answer

$\mu = 5 \pi 	imes 10^{-4}, \chi = 1249$

Vedclass · Answer

(C) The magnetic flux $\phi$ is given by $\phi = B \cdot A$, where $B$ is the magnetic field and $A$ is the cross-sectional area.$B = \frac{\phi}{A} = \frac{2.4 \pi 	imes 10^{-5} \ Wb}{2 	imes 10^{-5} \ m^2} = 1.2 \pi \ T$.Now, the relation between magnetic field $B$ and magnetising field $H$ is $B = \mu H$.$\mu = \frac{B}{H} = \frac{1.2 \pi \ T}{2400 \ A/m} = 5 \pi 	imes 10^{-4} \ T \cdot m/A$.The relative permeability $\mu_r$ is given by $\mu_r = \frac{\mu}{\mu_0}$.$\mu_r = \frac{5 \pi 	imes 10^{-4}}{4 \pi 	imes 10^{-7}} = 1250$.Since $\mu_r = 1 + \chi$, the magnetic susceptibility $\chi$ is $\chi = \mu_r - 1 = 1250 - 1 = 1249$.

Similar Questions

What is the magnetic susceptibility of a medium if its relative permeability is $0.85$?

An iron rod is placed parallel to a magnetic field of intensity $2000 \text{ A/m}$. The magnetic flux through the rod is $6 \times 10^{-4} \text{ Wb}$ and its cross-sectional area is $3 \text{ cm}^2$. The magnetic permeability of the rod in $\text{Wb/(A} \cdot \text{m)}$ is:

The magnetic susceptibility of the material of a rod is $599$. The absolute permeability of the material of the rod will be $\left[\mu_0=4 \pi \times 10^{-7} \text{ SI unit}\right]$.

If the magnetizing field on a ferromagnetic material is increased,its permeability

$A$ solenoid has a core of material with relative permeability $501$. The windings of the solenoid are insulated from the core and carry a current of $2.5 \ A$. If the number of turns is $900$ per metre,the magnetization in $A \ m^{-1}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module