Answer
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Hint: We have a solenoid of length $10cm$ having $100$ number of turns. A small magnet having a coercivity of $3 \times {10^3}A{m^{ - 1}}$. We have to find the current required to be passed through the solenoid so that the magnet will get demagnetized when it is placed inside the solenoid.
Complete step by step answer:
The particular intensity of the magnetic field at which the magnet is demagnetized is called the coercivity of the magnet.
The coercivity of the magnet can be written as,
$H = \dfrac{B}{{{\mu _0}}}$
For a solenoid, the magnetic field will be,
$B = {\mu _0}nI$
where $B$ is the magnetic field, n is the number of turns per unit length, ${\mu _0}$ is the permeability of free space, and $I$ is the current through each turn,
$\therefore \dfrac{B}{{{\mu _0}}} = H = nI$
The coercivity of the small magnet is given by
$H = 3 \times {10^3}A{m^{ - 1}}$
The length of the solenoid is given by,
$L = 10cm = 0.1m$
The number of turns of the solenoid is given by,
$N = 100$
The number of turns per unit length can be written as,
$n = \dfrac{{100}}{{0.1}}$
From this we get,
$I = \dfrac{H}{n}$
It is given that, $H = 3 \times {10^3}A{m^{ - 1}}$ and $n = \dfrac{{100}}{{0.1}}$
Substituting the values in the above equation, we get
$I = \dfrac{{3 \times {{10}^3}}}{{\dfrac{{100}}{{0.1}}}} = \dfrac{{300}}{{100}} = 3A$
Therefore the current required to be passed in a solenoid of length $10cm$ and number of turns $100$, so that the magnet gets demagnetized when inside the solenoid is $3A$.
Hence, the correct answer is option (C).
Note: A solenoid is a long closely wound helical coil. The magnetic moment per unit volume of a material is defined as the intensity of magnetization. Ferromagnetic substances are substances that are strongly magnetized in the presence of a magnetic field. The material will be magnetized in the direction of the magnetic field.
Complete step by step answer:
The particular intensity of the magnetic field at which the magnet is demagnetized is called the coercivity of the magnet.
The coercivity of the magnet can be written as,
$H = \dfrac{B}{{{\mu _0}}}$
For a solenoid, the magnetic field will be,
$B = {\mu _0}nI$
where $B$ is the magnetic field, n is the number of turns per unit length, ${\mu _0}$ is the permeability of free space, and $I$ is the current through each turn,
$\therefore \dfrac{B}{{{\mu _0}}} = H = nI$
The coercivity of the small magnet is given by
$H = 3 \times {10^3}A{m^{ - 1}}$
The length of the solenoid is given by,
$L = 10cm = 0.1m$
The number of turns of the solenoid is given by,
$N = 100$
The number of turns per unit length can be written as,
$n = \dfrac{{100}}{{0.1}}$
From this we get,
$I = \dfrac{H}{n}$
It is given that, $H = 3 \times {10^3}A{m^{ - 1}}$ and $n = \dfrac{{100}}{{0.1}}$
Substituting the values in the above equation, we get
$I = \dfrac{{3 \times {{10}^3}}}{{\dfrac{{100}}{{0.1}}}} = \dfrac{{300}}{{100}} = 3A$
Therefore the current required to be passed in a solenoid of length $10cm$ and number of turns $100$, so that the magnet gets demagnetized when inside the solenoid is $3A$.
Hence, the correct answer is option (C).
Note: A solenoid is a long closely wound helical coil. The magnetic moment per unit volume of a material is defined as the intensity of magnetization. Ferromagnetic substances are substances that are strongly magnetized in the presence of a magnetic field. The material will be magnetized in the direction of the magnetic field.
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