Question

A bar magnet has coercivity $4\times 10^{3}A/m$. It is desired to demagnetize it by inserting it inside a solenoid \[12cm\] long and has \[\text{60 turns}\]. The current that should be sent through the solenoid is:
\[\begin{align}
  & A.8A \\
 & B.10A \\
 & C.12A \\
 & D.14A \\
\end{align}\]

Answer 1

Hint: Coercivity of a magnet, like a ferromagnetic material is defined as the ability of the material to resist any external magnetic field without getting demagnetized i.e. losing its magnetic properties. Here, the magnet is demagnetized by keeping it inside a solenoid, and then the magnetic field produced by the solenoid will demagnetize the magnet.

Formula used:
$H=nI$

Complete step-by-step answer:
Coercivity of a magnetic material can also be defined as the intensity of the external magnetic field which can reduce the magnetic property of the given material to zero after reaching its saturation point.
We know that magnetic intensity is denoted by $H$ as is the measure of magnetic field strength, with SI units $A/m$.
The magnetic field due to a solenoid is given as $H=nI$ where, $I$ is current in the solenoid and $n$ is the number of winding s$N$ per unit length $l$ of the solenoid, or $n=\dfrac{N}{l}$.
Here it is given that coercivity of the bar magnet is $4\times 10^{3}A/m$, which means the intensity of the applied magnetic field is given as $H=4\times 10^{3}A/m$
Also given that the magnet is kept in a solenoid with $N=60turns$ and $l=12cm=0.12m$ ,then using the magnetic field due to a solenoid, we can say that the current in the solenoid $I$ is given as $I=\dfrac{H}{n}$
Substituting the values, we get $I=\dfrac{4\times 10^{3}}{\dfrac{60}{0.12}}=\dfrac{4\times10^{3}}{500}=8A$
Hence the current in the solenoid to demagnetize the magnet is \[8A\]

So, the correct answer is “Option A”.

Note: Magnetic field is a vector which gives the effect of a magnetic material and is denoted by $B$, with SI units $T$. However, in vacuum, $H=\dfrac{B}{\mu_{0}}$, where $\mu_{0}$ is the permeability of vacuum. Also $H$ and $B$ both indicate magnetic fields, but vary on how they are measured or obtained, hence have different definitions and SI units.