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A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 220 with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
- a)0.48G
- b)0.28G
- c)0.58G
- d)0.38G
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A magnetic needle free to rotate in a vertical plane parallel to the m...
We know the horizontal component of earth magnetic field H=R cos θ
Where θ=angle of dip=220 and the value of H=0.35G
So, putting the values respectively R= H/cosθ=0.35/cos22=0.35/0.927=0.377=0.38G
Where θ=angle of dip=220 and the value of H=0.35G
So, putting the values respectively R= H/cosθ=0.35/cos22=0.35/0.927=0.377=0.38G
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A magnetic needle free to rotate in a vertical plane parallel to the m...
To determine the magnitude of the Earth's magnetic field at a given place, we can use the information provided about the orientation of a magnetic needle in that location.
Given:
- The magnetic needle is free to rotate in a vertical plane parallel to the magnetic meridian.
- The north tip of the needle points down at an angle of 220° with the horizontal.
- The horizontal component of the Earth's magnetic field at the place is known to be 0.35 G.
To solve this problem, we can use the concept of the horizontal component of the Earth's magnetic field and the trigonometric relationship between the angle of deviation and the magnetic field.
Let's break down the solution into steps:
Step 1: Define the variables
Let:
- θ be the angle of deviation (220°)
- B be the magnitude of the Earth's magnetic field
- Bh be the horizontal component of the Earth's magnetic field (0.35 G)
Step 2: Find the vertical component of the Earth's magnetic field
The vertical component of the Earth's magnetic field (Bv) can be calculated using the formula:
Bv = Bh * tan(θ)
Substituting the given values:
Bv = 0.35 G * tan(220°)
Step 3: Use Pythagorean theorem to find the magnitude of the Earth's magnetic field
The magnitude of the Earth's magnetic field (B) can be found using the Pythagorean theorem:
B = sqrt(Bh^2 + Bv^2)
Substituting the given values:
B = sqrt((0.35 G)^2 + (0.35 G * tan(220°))^2)
B = sqrt(0.1225 G^2 + (0.35 G * (-0.7265))^2)
B = sqrt(0.1225 G^2 + 0.0904 G^2)
B = sqrt(0.2129 G^2)
B = 0.461 G
Step 4: Determine the correct option
The calculated magnitude of the Earth's magnetic field is approximately 0.461 G. Among the given options, the closest value is 0.38 G (option D).
Therefore, the correct answer is option D: 0.38 G.
Given:
- The magnetic needle is free to rotate in a vertical plane parallel to the magnetic meridian.
- The north tip of the needle points down at an angle of 220° with the horizontal.
- The horizontal component of the Earth's magnetic field at the place is known to be 0.35 G.
To solve this problem, we can use the concept of the horizontal component of the Earth's magnetic field and the trigonometric relationship between the angle of deviation and the magnetic field.
Let's break down the solution into steps:
Step 1: Define the variables
Let:
- θ be the angle of deviation (220°)
- B be the magnitude of the Earth's magnetic field
- Bh be the horizontal component of the Earth's magnetic field (0.35 G)
Step 2: Find the vertical component of the Earth's magnetic field
The vertical component of the Earth's magnetic field (Bv) can be calculated using the formula:
Bv = Bh * tan(θ)
Substituting the given values:
Bv = 0.35 G * tan(220°)
Step 3: Use Pythagorean theorem to find the magnitude of the Earth's magnetic field
The magnitude of the Earth's magnetic field (B) can be found using the Pythagorean theorem:
B = sqrt(Bh^2 + Bv^2)
Substituting the given values:
B = sqrt((0.35 G)^2 + (0.35 G * tan(220°))^2)
B = sqrt(0.1225 G^2 + (0.35 G * (-0.7265))^2)
B = sqrt(0.1225 G^2 + 0.0904 G^2)
B = sqrt(0.2129 G^2)
B = 0.461 G
Step 4: Determine the correct option
The calculated magnitude of the Earth's magnetic field is approximately 0.461 G. Among the given options, the closest value is 0.38 G (option D).
Therefore, the correct answer is option D: 0.38 G.
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A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at220with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.a)0.48Gb)0.28Gc)0.58Gd)0.38GCorrect answer is option 'D'. Can you explain this answer?
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A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at220with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.a)0.48Gb)0.28Gc)0.58Gd)0.38GCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at220with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.a)0.48Gb)0.28Gc)0.58Gd)0.38GCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at220with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.a)0.48Gb)0.28Gc)0.58Gd)0.38GCorrect answer is option 'D'. Can you explain this answer?.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at220with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.a)0.48Gb)0.28Gc)0.58Gd)0.38GCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at220with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.a)0.48Gb)0.28Gc)0.58Gd)0.38GCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at220with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.a)0.48Gb)0.28Gc)0.58Gd)0.38GCorrect answer is option 'D'. Can you explain this answer?.
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ample number of questions to practice A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at220with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.a)0.48Gb)0.28Gc)0.58Gd)0.38GCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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