Magnetics Ques 104
- A bar magnet is demagnetised by inserting it inside a solenoid of length $0.2$ $m$, $100$ turns and carrying a current of $5.2$ A. The coercivity of the bar magnet is
(Main 2019, 9 Jan I)
(a) $1200$ $ A / m$
(b) $285$ $ A / m$
(c) $2600$ $ A / m$
(d) $520$ $ A / m$
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Answer:
Correct Answer: 104.(c)
Solution:
- Coercivity of a bar magnet is the value of magnetic field intensity $(H)$ that is needed to reduce magnetisation to zero. Since, for a solenoid magnetic induction is given as,
$ B=\mu _0 n I $ $\quad$ …….(i)
where, $n$ is the number of turns $(N)$ per unit, length $(l)$ and $I$ is the current.
Also,
$ B=\mu _0 H $ $\quad$ …….(ii)
$\therefore$ From Eqs. (i) and (ii), we get
$ \mu _0 n I=\mu _0 H \text { or } H=n I=\frac{N}{l} I $
Substituting the given values, we get
$ H=\frac{100}{0.2} \times 5.2=2600 A / m $
Thus, the value of coercivity of the bar magnet is $2600$ $A / m$.
BETA
