A bar magnet has a coercivity of $4 \times 10^{3} A m^{- 1}$ . It is placed inside a solenoid of length $12 c m$ having 60 turns in order to demagnetize it, the amount of current that should be passed through the solenoid is:

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Answer 1

Given: $H = 4 \times 1 0^{3} A m^{- 1}$

$L = 12 c m$

$N = 60$

Bar magnet requires a magnetic intensity ( $H = 4 \times 1 0^{3} A m^{- 1}$ ) to become demagnetized.

Let $n$ be the number of turns per unit length.

$\Rightarrow n = \frac{N}{L} = \frac{6 0}{0 . 1 2} = 500 t u r n s / m e t r e$

Now, $H = n I$

$\Rightarrow I = \frac{H}{n} = \frac{4 \times 1 0 ^{3}}{5 0 0} = 8 A$

Answer 2

Given: $H = 4 \times 1 0^{3} A m^{- 1}$

$L = 12 c m$

$N = 60$

Bar magnet requires a magnetic intensity ( $H = 4 \times 1 0^{3} A m^{- 1}$ ) to become demagnetized.

Let $n$ be the number of turns per unit length.

$\Rightarrow n = \frac{N}{L} = \frac{6 0}{0 . 1 2} = 500 t u r n s / m e t r e$

Now, $H = n I$

$\Rightarrow I = \frac{H}{n} = \frac{4 \times 1 0 ^{3}}{5 0 0} = 8 A$