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Question: A bar magnet has coercivity \(4\times 10^{3}A/m\). It is desired to demagnetize it by inserting it i...

A bar magnet has coercivity 4×103A/m4\times 10^{3}A/m. It is desired to demagnetize it by inserting it inside a solenoid 12cm12cm long and has 60 turns\text{60 turns}. The current that should be sent through the solenoid is:

& A.8A \\\ & B.10A \\\ & C.12A \\\ & D.14A \\\ \end{aligned}$$
Explanation

Solution

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=nIH=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 HH as is the measure of magnetic field strength, with SI units A/mA/m.
The magnetic field due to a solenoid is given as H=nIH=nI where, II is current in the solenoid and nn is the number of winding sNN per unit length ll of the solenoid, or n=Nln=\dfrac{N}{l}.
Here it is given that coercivity of the bar magnet is 4×103A/m4\times 10^{3}A/m, which means the intensity of the applied magnetic field is given as H=4×103A/mH=4\times 10^{3}A/m
Also given that the magnet is kept in a solenoid with N=60turnsN=60turns and l=12cm=0.12ml=12cm=0.12m ,then using the magnetic field due to a solenoid, we can say that the current in the solenoid II is given as I=HnI=\dfrac{H}{n}
Substituting the values, we get I=4×103600.12=4×103500=8AI=\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 8A8A

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 BB, with SI units TT. However, in vacuum, H=Bμ0H=\dfrac{B}{\mu_{0}}, where μ0\mu_{0} is the permeability of vacuum. Also HH and BB both indicate magnetic fields, but vary on how they are measured or obtained, hence have different definitions and SI units.