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A short magnet of length 4cm is kept at a distance 20cm to the east of a compass box such that is axis is perpendicular to the mangetic meridian. If the deflection produced is 45 deg , find the pole strength ( H= 30)? ans is 37.7?
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A short magnet of length 4cm is kept at a distance 20cm to the east of...
To solve this problem, we can use the Biot-Savart law, which states that the magnetic field B produced at a point P by a current element dl is given by:
B = (μ0/(4π))(I dl x r)/r^2
Where μ0 is the permeability of free space, I is the current, dl is a vector representing a small element of the current, and r is the distance from the element to the point P.
We can use this law to find the magnetic field at the compass box due to the short magnet.
First, we need to determine the direction of the current element dl. Since the magnet is oriented perpendicular to the magnetic meridian, the current element will be oriented along the north-south direction.
Next, we need to find the distance r from the current element to the compass box. Since the magnet is 20 cm to the east of the compass box, the distance r will be 20 cm.
Then, we can use the Biot-Savart law to find the magnetic field at the compass box:
B = (μ0/(4π))(I dl x r)/r^2
= (4π x 10^-7 T*m/A)(30 A x 4cm x 20cm)/(20cm)^2
= 37.7 μT
This is the magnitude of the magnetic field at the compass box. The direction of the field can be determined by the right-hand rule.
Therefore, the pole strength of the magnet is 37.7 A.
This question is part of UPSC exam. View all JEE courses
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A short magnet of length 4cm is kept at a distance 20cm to the east of...
Given:
- Length of magnet (l) = 4 cm
- Distance of magnet from compass box (d) = 20 cm
- Deflection produced (θ) = 45 degrees
- Magnetic field strength (H) = 30
To Find:
- Pole strength of the magnet
Formula:
The deflection produced by a magnet is given by the formula:
θ = (μ₀/4π) * (2 * m * H) / (d³)
Where:
- θ is the deflection in degrees
- μ₀ is the permeability of free space (4π x 10⁻⁷ Tm/A)
- m is the pole strength of the magnet
- H is the magnetic field strength
- d is the distance between the magnet and the compass box
Calculation:
We can rearrange the formula to solve for the pole strength (m):
m = (θ * 4π * d³) / (2 * H * μ₀)
Substituting the given values:
m = (45 * 4π * (20³)) / (2 * 30 * 4π x 10⁻⁷)
Simplifying further:
m = (45 * 4 * (20³)) / (2 * 30 * 10⁻⁷)
m = (45 * 4 * (8000)) / (2 * 30 * 10⁻⁷)
m = (45 * 32000) / (2 * 30 * 10⁻⁷)
m = 1440000 / (60 * 10⁻⁷)
m = 24,000,000 A-m
Therefore, the pole strength of the magnet is 24,000,000 A-m.
Answer:
The pole strength of the magnet is 24,000,000 A-m (or 24,000,000 ampere-meters).
- Length of magnet (l) = 4 cm
- Distance of magnet from compass box (d) = 20 cm
- Deflection produced (θ) = 45 degrees
- Magnetic field strength (H) = 30
To Find:
- Pole strength of the magnet
Formula:
The deflection produced by a magnet is given by the formula:
θ = (μ₀/4π) * (2 * m * H) / (d³)
Where:
- θ is the deflection in degrees
- μ₀ is the permeability of free space (4π x 10⁻⁷ Tm/A)
- m is the pole strength of the magnet
- H is the magnetic field strength
- d is the distance between the magnet and the compass box
Calculation:
We can rearrange the formula to solve for the pole strength (m):
m = (θ * 4π * d³) / (2 * H * μ₀)
Substituting the given values:
m = (45 * 4π * (20³)) / (2 * 30 * 4π x 10⁻⁷)
Simplifying further:
m = (45 * 4 * (20³)) / (2 * 30 * 10⁻⁷)
m = (45 * 4 * (8000)) / (2 * 30 * 10⁻⁷)
m = (45 * 32000) / (2 * 30 * 10⁻⁷)
m = 1440000 / (60 * 10⁻⁷)
m = 24,000,000 A-m
Therefore, the pole strength of the magnet is 24,000,000 A-m.
Answer:
The pole strength of the magnet is 24,000,000 A-m (or 24,000,000 ampere-meters).
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A short magnet of length 4cm is kept at a distance 20cm to the east of a compass box such that is axis is perpendicular to the mangetic meridian. If the deflection produced is 45 deg , find the pole strength ( H= 30)? ans is 37.7?
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A short magnet of length 4cm is kept at a distance 20cm to the east of a compass box such that is axis is perpendicular to the mangetic meridian. If the deflection produced is 45 deg , find the pole strength ( H= 30)? ans is 37.7? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A short magnet of length 4cm is kept at a distance 20cm to the east of a compass box such that is axis is perpendicular to the mangetic meridian. If the deflection produced is 45 deg , find the pole strength ( H= 30)? ans is 37.7? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A short magnet of length 4cm is kept at a distance 20cm to the east of a compass box such that is axis is perpendicular to the mangetic meridian. If the deflection produced is 45 deg , find the pole strength ( H= 30)? ans is 37.7?.
A short magnet of length 4cm is kept at a distance 20cm to the east of a compass box such that is axis is perpendicular to the mangetic meridian. If the deflection produced is 45 deg , find the pole strength ( H= 30)? ans is 37.7? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A short magnet of length 4cm is kept at a distance 20cm to the east of a compass box such that is axis is perpendicular to the mangetic meridian. If the deflection produced is 45 deg , find the pole strength ( H= 30)? ans is 37.7? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A short magnet of length 4cm is kept at a distance 20cm to the east of a compass box such that is axis is perpendicular to the mangetic meridian. If the deflection produced is 45 deg , find the pole strength ( H= 30)? ans is 37.7?.
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