A bar magnet is demagnetized by inserting it inside a solenoid of length $0.2m$, $100$ turns, and carrying a current of $5.2{\text{A}}{\text{.}}$ The coercivity of the bar magnet is:
(A) $1200{\text{ A}}/m$
(B) $2600{\text{ A}}/m$
(C) $5200{\text{ A}}/m$
(D) $5285{\text{ A}}/m$
Answer
638.7k+ views
Hint
To solve this question, we need to find the intensity of the magnetic field inside the solenoid. Since the bar magnet is demagnetized by placing it inside the given solenoid, so the magnetic field intensity of the solenoid will be the required coercivity.
$B = \dfrac{{{\mu _0}Ni}}{l}$, where $B$ is the magnetic field inside a solenoid of length $l$ having $N$ number of turns and carrying a current $i$.
$H = \dfrac{B}{{{\mu _0}}}$, where $H$ is the magnetic field intensity and $B$ is the magnetic field strength.
Complete step by step answer
We know that the magnetic field inside a solenoid is given by
$B = \dfrac{{{\mu _0}Ni}}{l}$
According to the question
$N = 100$
$l = 0.2m$
$i = 5.2A$
Substituting these in the above equation, we have
$B = \dfrac{{{\mu _0}(100)(5.2)}}{{0.2}}$
Or $B = 2600{\mu _0}$ (1)
Now, we know that the intensity of magnetic field is given by
$H = \dfrac{B}{{{\mu _0}}}$
Substituting $B$ from (1), we get
$H = \dfrac{{2600{\mu _0}}}{{{\mu _0}}}$
$H = 2600{\text{ A/}}m$
This is the magnetic field intensity inside the solenoid.
Now, the coercivity of a material is the magnetic field intensity required to demagnetize it. Since the bar magnet is demagnetized by placing it in the solenoid, so the coercivity of the bar magnet will be equal to the intensity of the solenoid, i.e. $2600{\text{ A/}}m$
Hence, the correct answer is option (B), $2600{\text{ A/}}m$.
Note
The term ‘coercivity’ is basically related to the hysteresis loop, which is a plot between the intensity and the magnetic field intensity and the intensity of magnetization. The hysteresis effect has many applications in ferromagnetic materials including credit cards, radio tape, hard disks etc. which require retaining some memory. Also, it has applications in the devices such as transformers, permanent magnets, and electromagnets etc. which are required to retain the magnetic intensity. For such applications, materials having high coercivity, known as hard magnets are required.
To solve this question, we need to find the intensity of the magnetic field inside the solenoid. Since the bar magnet is demagnetized by placing it inside the given solenoid, so the magnetic field intensity of the solenoid will be the required coercivity.
$B = \dfrac{{{\mu _0}Ni}}{l}$, where $B$ is the magnetic field inside a solenoid of length $l$ having $N$ number of turns and carrying a current $i$.
$H = \dfrac{B}{{{\mu _0}}}$, where $H$ is the magnetic field intensity and $B$ is the magnetic field strength.
Complete step by step answer
We know that the magnetic field inside a solenoid is given by
$B = \dfrac{{{\mu _0}Ni}}{l}$
According to the question
$N = 100$
$l = 0.2m$
$i = 5.2A$
Substituting these in the above equation, we have
$B = \dfrac{{{\mu _0}(100)(5.2)}}{{0.2}}$
Or $B = 2600{\mu _0}$ (1)
Now, we know that the intensity of magnetic field is given by
$H = \dfrac{B}{{{\mu _0}}}$
Substituting $B$ from (1), we get
$H = \dfrac{{2600{\mu _0}}}{{{\mu _0}}}$
$H = 2600{\text{ A/}}m$
This is the magnetic field intensity inside the solenoid.
Now, the coercivity of a material is the magnetic field intensity required to demagnetize it. Since the bar magnet is demagnetized by placing it in the solenoid, so the coercivity of the bar magnet will be equal to the intensity of the solenoid, i.e. $2600{\text{ A/}}m$
Hence, the correct answer is option (B), $2600{\text{ A/}}m$.
Note
The term ‘coercivity’ is basically related to the hysteresis loop, which is a plot between the intensity and the magnetic field intensity and the intensity of magnetization. The hysteresis effect has many applications in ferromagnetic materials including credit cards, radio tape, hard disks etc. which require retaining some memory. Also, it has applications in the devices such as transformers, permanent magnets, and electromagnets etc. which are required to retain the magnetic intensity. For such applications, materials having high coercivity, known as hard magnets are required.
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