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Magnetic Properties of Matter question

2019 · 9 Jan · Shift 1 · Q46
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Magnetic Properties of Matter question

2019 · 9 Jan · Shift 1 · Q46

JEE MainPhysicsMagnetic Properties of MatterMCQ+4 / −1
A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. The corecivity of the bar magnet is :
  1. A
    285 A/m
  2. B
    2600 A/m
  3. C
    520 A/m
  4. D
    1200 A/m
View written solutionFree

Correct answer: B

  1. Concept used

To demagnetize a bar magnet, the magnetizing field produced by the solenoid must be equal to the coercivity HcH_cHc​ of the magnet.

For a solenoid, the magnetic field intensity is

H=nI=NLIH = nI = \frac{N}{L} IH=nI=LN​I

where:

  • N=100N = 100N=100 turns
  • L=0.2 mL = 0.2\,\text{m}L=0.2m
  • I=5.2 AI = 5.2\,\text{A}I=5.2A
  1. Calculate turns per unit length
n=NL=1000.2=500 turns/mn = \frac{N}{L} = \frac{100}{0.2} = 500\,\text{turns/m}n=LN​=0.2100​=500turns/m
  1. Calculate magnetic field intensity
H=nI=500×5.2=2600 A/mH = nI = 500 \times 5.2 = 2600\,\text{A/m}H=nI=500×5.2=2600A/m
  1. Interpretation

This field intensity is the coercivity required to demagnetize the bar magnet.

So,

Hc=2600 A/mH_c = 2600\,\text{A/m}Hc​=2600A/m
  1. Check options
  • A: 285 A/m285\,\text{A/m}285A/m ❌
  • B: 2600 A/m2600\,\text{A/m}2600A/m ✅
  • C: 520 A/m520\,\text{A/m}520A/m ❌
  • D: 1200 A/m1200\,\text{A/m}1200A/m ❌

Therefore, the correct option is B.

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