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An iron rod of volume 10-4 m3 & relative permeability 1000 is placed inside a long solenoid wound with 5 turns per cm. If a current of 0.5 A is passed through the solenoid, find the magnetic moment of the rod.?
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An iron rod of volume 10-4 m3 & relative permeability 1000 is placed i...
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An iron rod of volume 10-4 m3 & relative permeability 1000 is placed i...
An iron rod with a volume of 10^-4 m^3 would be a very small rod. It would have dimensions that depend on its shape, but if we assume it is a cylinder, we can calculate its approximate length and diameter.
The volume of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height. Since we are given the volume V, we can rearrange the formula to solve for the radius:
V = πr^2h
10^-4 = πr^2h
Assuming the height h is equal to the length of the rod, we can substitute h with L:
10^-4 = πr^2L
We can rearrange the formula to solve for the radius r:
r^2 = (10^-4) / (πL)
r = √[(10^-4) / (πL)]
Without knowing the specific length L of the rod, we cannot calculate the exact radius. However, we can make some assumptions to estimate the dimensions.
Let's assume the length L of the rod is 0.1 m (10 cm), which is a reasonable size for an iron rod. Substituting this value into the formula, we can calculate the radius:
r = √[(10^-4) / (π*0.1)]
r ≈ √[(10^-4) / (0.314)]
r ≈ √[(3.18 x 10^-4)]
r ≈ 0.0178 m (approximately 1.78 cm)
Therefore, for an iron rod with a volume of 10^-4 m^3, assuming it is a cylinder with a length of 0.1 m, the approximate radius would be 0.0178 m (approximately 1.78 cm).
The volume of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height. Since we are given the volume V, we can rearrange the formula to solve for the radius:
V = πr^2h
10^-4 = πr^2h
Assuming the height h is equal to the length of the rod, we can substitute h with L:
10^-4 = πr^2L
We can rearrange the formula to solve for the radius r:
r^2 = (10^-4) / (πL)
r = √[(10^-4) / (πL)]
Without knowing the specific length L of the rod, we cannot calculate the exact radius. However, we can make some assumptions to estimate the dimensions.
Let's assume the length L of the rod is 0.1 m (10 cm), which is a reasonable size for an iron rod. Substituting this value into the formula, we can calculate the radius:
r = √[(10^-4) / (π*0.1)]
r ≈ √[(10^-4) / (0.314)]
r ≈ √[(3.18 x 10^-4)]
r ≈ 0.0178 m (approximately 1.78 cm)
Therefore, for an iron rod with a volume of 10^-4 m^3, assuming it is a cylinder with a length of 0.1 m, the approximate radius would be 0.0178 m (approximately 1.78 cm).
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An iron rod of volume 10-4 m3 & relative permeability 1000 is placed inside a long solenoid wound with 5 turns per cm. If a current of 0.5 A is passed through the solenoid, find the magnetic moment of the rod.?
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An iron rod of volume 10-4 m3 & relative permeability 1000 is placed inside a long solenoid wound with 5 turns per cm. If a current of 0.5 A is passed through the solenoid, find the magnetic moment of the rod.? 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 10-4 m3 & relative permeability 1000 is placed inside a long solenoid wound with 5 turns per cm. If a current of 0.5 A is passed through the solenoid, find the magnetic moment of the rod.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An iron rod of volume 10-4 m3 & relative permeability 1000 is placed inside a long solenoid wound with 5 turns per cm. If a current of 0.5 A is passed through the solenoid, find the magnetic moment of the rod.?.
An iron rod of volume 10-4 m3 & relative permeability 1000 is placed inside a long solenoid wound with 5 turns per cm. If a current of 0.5 A is passed through the solenoid, find the magnetic moment of the rod.? 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 10-4 m3 & relative permeability 1000 is placed inside a long solenoid wound with 5 turns per cm. If a current of 0.5 A is passed through the solenoid, find the magnetic moment of the rod.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An iron rod of volume 10-4 m3 & relative permeability 1000 is placed inside a long solenoid wound with 5 turns per cm. If a current of 0.5 A is passed through the solenoid, find the magnetic moment of the rod.?.
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