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A 10 cm long bar magnet of magnetic moment 1.34 Am2 is placed in the magnetic meridian with its south pole pointing geographical south. The neutral point is obtained at a distance of 15 cm from the center of the magnet. Calculate the horizontal component of earth’s magnetic field.
  • a)
    0.12 × 10–4 T
  • b)
    0.21× 10–4 T
  • c)
    0.34 × 10–4 T
  • d)
    0.87 × 10–7 T
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A 10 cm long bar magnet of magnetic moment 1.34 Am2 is placed in the ...
As the magnet is placed with its south pole pointing south, hence the neutral point lies on the equatorial line. At the neutral point, the magnetic field B due to the magnet becomes equal and opposite to the horizontal component of earth’s magnetic field i.e., BH.
Hence, if M be magnetic dipole moment of the magnet of length 2l and r the distance of the neutral point from its center, then
Given that µ0 = 4π × 10–7 T mA–1, M = 1.34 Am2,
r = 15 cm = 0.15 m and l = 5.0 cm = 0.05 m
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Most Upvoted Answer
A 10 cm long bar magnet of magnetic moment 1.34 Am2 is placed in the ...
To solve this problem, we need to understand the concept of a neutral point and how it relates to the magnetic field of the Earth.

1. Understanding the Neutral Point:
- A neutral point is a point where the magnetic field due to a bar magnet is balanced by the magnetic field of the Earth.
- At the neutral point, the magnetic field due to the bar magnet cancels out the horizontal component of the Earth's magnetic field.

2. Given Data:
- Length of the magnet (l) = 10 cm = 0.1 m
- Magnetic moment of the magnet (m) = 1.34 Am²
- Distance of the neutral point from the center of the magnet (d) = 15 cm = 0.15 m

3. Calculating the Earth's Magnetic Field:
- The magnetic field due to a bar magnet at a point on the axial line at a distance (d) is given by the formula:
B = (μ₀/4π) * (2m/d³) [1]
where μ₀ is the permeability of free space (4π × 10⁻⁷ Tm/A).

- At the neutral point, the magnetic field due to the bar magnet (B₁) is equal in magnitude and opposite in direction to the horizontal component of the Earth's magnetic field (B₂).
Therefore, B₁ = B₂ [2]

- The horizontal component of the Earth's magnetic field (B₂) can be calculated using equations [1] and [2].
Equating B₁ and B₂:
(μ₀/4π) * (2m/d³) = B₂
B₂ = (μ₀/4π) * (2m/d³) [3]

4. Substituting the Given Values:
- Substituting the given values into equation [3]:
B₂ = (4π × 10⁻⁷ Tm/A) / (4π) * (2 × 1.34 Am²) / (0.15 m)³
≈ 0.34 × 10⁻⁴ T

5. Answer:
The horizontal component of the Earth's magnetic field is approximately 0.34 × 10⁻⁴ T.
Therefore, the correct answer is option C.
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A 10 cm long bar magnet of magnetic moment 1.34 Am2 is placed in the magnetic meridian with its south pole pointing geographical south. The neutral point is obtained at a distance of 15 cm from the center of the magnet. Calculate the horizontal component of earth’s magnetic field.a)0.12 × 10–4 Tb)0.21× 10–4 Tc)0.34 × 10–4 Td)0.87 × 10–7 TCorrect answer is option 'C'. Can you explain this answer?
Question Description
A 10 cm long bar magnet of magnetic moment 1.34 Am2 is placed in the magnetic meridian with its south pole pointing geographical south. The neutral point is obtained at a distance of 15 cm from the center of the magnet. Calculate the horizontal component of earth’s magnetic field.a)0.12 × 10–4 Tb)0.21× 10–4 Tc)0.34 × 10–4 Td)0.87 × 10–7 TCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A 10 cm long bar magnet of magnetic moment 1.34 Am2 is placed in the magnetic meridian with its south pole pointing geographical south. The neutral point is obtained at a distance of 15 cm from the center of the magnet. Calculate the horizontal component of earth’s magnetic field.a)0.12 × 10–4 Tb)0.21× 10–4 Tc)0.34 × 10–4 Td)0.87 × 10–7 TCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 cm long bar magnet of magnetic moment 1.34 Am2 is placed in the magnetic meridian with its south pole pointing geographical south. The neutral point is obtained at a distance of 15 cm from the center of the magnet. Calculate the horizontal component of earth’s magnetic field.a)0.12 × 10–4 Tb)0.21× 10–4 Tc)0.34 × 10–4 Td)0.87 × 10–7 TCorrect answer is option 'C'. Can you explain this answer?.
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