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A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is
(1 G = 10-4 T)
(1 G = 10-4 T)
- a)288.0 JT-1
- b)2.880 JT-1
- c)6.880 JT-1
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A bar magnet of length 14 cm is placed in the magnetic meridian with i...
CONCEPT:
We know that a neutral point is a point where the earth's magnetic field's horizontal component cancels the magnetic field due to the bar.
We know that a neutral point is a point where the earth's magnetic field's horizontal component cancels the magnetic field due to the bar.
CALCULATION:
Given:
The magnetic field due to bar magnet at the neutral point N is written as;
So at N,
Bm = BH
⇒ M =2.880 J/T
Hence option 2) is the correct answer.
Given:
The magnetic field due to bar magnet at the neutral point N is written as;
So at N,
Bm = BH
⇒ M =2.880 J/T
Hence option 2) is the correct answer.
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Community Answer
A bar magnet of length 14 cm is placed in the magnetic meridian with i...
Given information:
- Length of the bar magnet, l = 14 cm
- Distance to the neutral point, x = 18 cm
- Magnetic field at the neutral point, BH = 0.4 G
We need to find the magnetic moment of the magnet.
The formula for the magnetic field at a point on the axial line of a bar magnet is given by:
BH = μ0/4π * (2M * l) / (x^2 + l^2)^(3/2)
Where,
BH is the magnetic field at the neutral point
μ0 is the permeability of free space (4π × 10^-7 T m/A)
M is the magnetic moment of the magnet
l is the length of the magnet
x is the distance to the neutral point
Let's substitute the given values into the formula and solve for M:
0.4 = (4π × 10^-7 T m/A) / (4π) * (2M * 0.14 m) / (0.18^2 + 0.14^2)^(3/2)
Simplifying the equation:
0.4 = (10^-7 T m/A) * M * 0.28 m / (0.0324 + 0.0196)^(3/2)
0.4 = (10^-7 T m/A) * M * 0.28 m / (0.0520)^(3/2)
0.4 = (10^-7 T m/A) * M * 0.28 m / (0.0520)^(3/2)
0.4 = (10^-7 T m/A) * M * 0.28 m / (0.0520)^(3/2)
0.4 = (10^-7 T m/A) * M * 0.28 m / (0.02704)
0.4 = (10^-7 T m/A) * M * 10.34 m
Dividing both sides by (10^-7 T m/A * 10.34 m):
0.4 / (10^-7 T m/A * 10.34 m) = M
M = 0.4 / (10^-7 * 10.34) A m^2
M ≈ 3.872 × 10^6 A m^2
Converting A m^2 to JT^-1:
M = 3.872 × 10^6 JT^-1
Therefore, the magnetic moment of the magnet is approximately 3.872 × 10^6 JT^-1, which is equivalent to option B.
- Length of the bar magnet, l = 14 cm
- Distance to the neutral point, x = 18 cm
- Magnetic field at the neutral point, BH = 0.4 G
We need to find the magnetic moment of the magnet.
The formula for the magnetic field at a point on the axial line of a bar magnet is given by:
BH = μ0/4π * (2M * l) / (x^2 + l^2)^(3/2)
Where,
BH is the magnetic field at the neutral point
μ0 is the permeability of free space (4π × 10^-7 T m/A)
M is the magnetic moment of the magnet
l is the length of the magnet
x is the distance to the neutral point
Let's substitute the given values into the formula and solve for M:
0.4 = (4π × 10^-7 T m/A) / (4π) * (2M * 0.14 m) / (0.18^2 + 0.14^2)^(3/2)
Simplifying the equation:
0.4 = (10^-7 T m/A) * M * 0.28 m / (0.0324 + 0.0196)^(3/2)
0.4 = (10^-7 T m/A) * M * 0.28 m / (0.0520)^(3/2)
0.4 = (10^-7 T m/A) * M * 0.28 m / (0.0520)^(3/2)
0.4 = (10^-7 T m/A) * M * 0.28 m / (0.0520)^(3/2)
0.4 = (10^-7 T m/A) * M * 0.28 m / (0.02704)
0.4 = (10^-7 T m/A) * M * 10.34 m
Dividing both sides by (10^-7 T m/A * 10.34 m):
0.4 / (10^-7 T m/A * 10.34 m) = M
M = 0.4 / (10^-7 * 10.34) A m^2
M ≈ 3.872 × 10^6 A m^2
Converting A m^2 to JT^-1:
M = 3.872 × 10^6 JT^-1
Therefore, the magnetic moment of the magnet is approximately 3.872 × 10^6 JT^-1, which is equivalent to option B.
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A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is(1 G = 10-4 T)a)288.0 JT-1b)2.880 JT-1c)6.880 JT-1d)None of theseCorrect answer is option 'B'. Can you explain this answer?
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A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is(1 G = 10-4 T)a)288.0 JT-1b)2.880 JT-1c)6.880 JT-1d)None of theseCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is(1 G = 10-4 T)a)288.0 JT-1b)2.880 JT-1c)6.880 JT-1d)None of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is(1 G = 10-4 T)a)288.0 JT-1b)2.880 JT-1c)6.880 JT-1d)None of theseCorrect answer is option 'B'. Can you explain this answer?.
A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is(1 G = 10-4 T)a)288.0 JT-1b)2.880 JT-1c)6.880 JT-1d)None of theseCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is(1 G = 10-4 T)a)288.0 JT-1b)2.880 JT-1c)6.880 JT-1d)None of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is(1 G = 10-4 T)a)288.0 JT-1b)2.880 JT-1c)6.880 JT-1d)None of theseCorrect answer is option 'B'. Can you explain this answer?.
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A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is(1 G = 10-4 T)a)288.0 JT-1b)2.880 JT-1c)6.880 JT-1d)None of theseCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is(1 G = 10-4 T)a)288.0 JT-1b)2.880 JT-1c)6.880 JT-1d)None of theseCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4 G, the magnetic moment of the magnet is(1 G = 10-4 T)a)288.0 JT-1b)2.880 JT-1c)6.880 JT-1d)None of theseCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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