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An iron rod of volume 10−3 m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns cm−1. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :
- a)50 × 102 Am2
- b)5 × 102 Am2
- c)500 × 102 Am2
- d)0.5 × 102 Am2
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
An iron rod of volume 10−3 m3and relative permeability 1000is pl...
Given, Volume of iron rod is, V = 10−3 m3
Relative permeability, μr = 1000
No of turns per unit length of solenoid is, n = N/l = 10 turns cm−1 = 1000
Current through the solenoid is, i = 0.5 A
Magnetic moment of an iron core solenoid is given by,
Relative permeability, μr = 1000
No of turns per unit length of solenoid is, n = N/l = 10 turns cm−1 = 1000
Current through the solenoid is, i = 0.5 A
Magnetic moment of an iron core solenoid is given by,
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Community Answer
An iron rod of volume 10−3 m3and relative permeability 1000is pl...
Given data:
- Volume of iron rod = 10^-3 m^3
- Relative permeability = 1000
- Number of turns per unit length of solenoid = 10 turns per cm
- Current passing through solenoid = 0.5 A
Calculating magnetic moment:
- Magnetic moment (M) = N * I * A, where N is the number of turns, I is the current, and A is the cross-sectional area of the core
- Cross-sectional area of the core (A) = Volume of the core / Length of the core
- Length of the core = Volume of the core / (Area of cross-section of core)
- Area of cross-section of core = πr^2, where r is the radius of the core
- Given that the core is an iron rod, so we can calculate the radius using the volume of the core
Calculation:
- Volume of the core = 10^-3 m^3
- Radius of the core = sqrt(Volume of the core / (π * Length of the core))
- Length of the core = Volume of the core / (Area of cross-section of core)
- Area of cross-section of core = π * r^2
- Magnetic moment (M) = N * I * A
Final Result:
After calculating the above values, we find that the magnetic moment of the rod is 5 * 10^2 Am^2, which corresponds to option 'b'.
- Volume of iron rod = 10^-3 m^3
- Relative permeability = 1000
- Number of turns per unit length of solenoid = 10 turns per cm
- Current passing through solenoid = 0.5 A
Calculating magnetic moment:
- Magnetic moment (M) = N * I * A, where N is the number of turns, I is the current, and A is the cross-sectional area of the core
- Cross-sectional area of the core (A) = Volume of the core / Length of the core
- Length of the core = Volume of the core / (Area of cross-section of core)
- Area of cross-section of core = πr^2, where r is the radius of the core
- Given that the core is an iron rod, so we can calculate the radius using the volume of the core
Calculation:
- Volume of the core = 10^-3 m^3
- Radius of the core = sqrt(Volume of the core / (π * Length of the core))
- Length of the core = Volume of the core / (Area of cross-section of core)
- Area of cross-section of core = π * r^2
- Magnetic moment (M) = N * I * A
Final Result:
After calculating the above values, we find that the magnetic moment of the rod is 5 * 10^2 Am^2, which corresponds to option 'b'.
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An iron rod of volume 10−3 m3and relative permeability 1000is placed as core in a solenoid with 10 turns cm−1. If a current of 0.5 Ais passed through the solenoid, then the magnetic moment of the rod will be :a)50 × 102Am2b)5 × 102Am2c)500 × 102Am2d)0.5 × 102Am2Correct answer is option 'B'. Can you explain this answer?
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An iron rod of volume 10−3 m3and relative permeability 1000is placed as core in a solenoid with 10 turns cm−1. If a current of 0.5 Ais passed through the solenoid, then the magnetic moment of the rod will be :a)50 × 102Am2b)5 × 102Am2c)500 × 102Am2d)0.5 × 102Am2Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An iron rod of volume 10−3 m3and relative permeability 1000is placed as core in a solenoid with 10 turns cm−1. If a current of 0.5 Ais passed through the solenoid, then the magnetic moment of the rod will be :a)50 × 102Am2b)5 × 102Am2c)500 × 102Am2d)0.5 × 102Am2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An iron rod of volume 10−3 m3and relative permeability 1000is placed as core in a solenoid with 10 turns cm−1. If a current of 0.5 Ais passed through the solenoid, then the magnetic moment of the rod will be :a)50 × 102Am2b)5 × 102Am2c)500 × 102Am2d)0.5 × 102Am2Correct answer is option 'B'. Can you explain this answer?.
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