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a. $30mA$

b. $60mA$

c. $3A$

d. $6A$

Answer

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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 $3A$.

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