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The SI unit of magnetic permebility is...

The `SI` unit of magnetic permebility is

A

`Wbm^-2A^-1`

B

`Wbm^-1A`

C

`Wbm^-1A^-1`

D

`WbmA^-1`

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The correct Answer is:
To determine the SI unit of magnetic permeability, we can start from the relationship between magnetic field (B), magnetic permeability (μ), current (I), and the geometry of the system. ### Step-by-Step Solution: 1. **Understanding the Formula**: We know that the magnetic field (B) can be expressed in terms of magnetic permeability (μ) and current (I). The formula is: \[ B = \frac{\mu}{4\pi} \cdot \frac{I \cdot dl \cdot \sin \theta}{r^2} \] where: - \( B \) is the magnetic field, - \( \mu \) is the magnetic permeability, - \( I \) is the current, - \( dl \) is a differential length element, - \( \theta \) is the angle between the current element and the line connecting the element to the point where the field is being measured, - \( r \) is the distance from the current element to the point where the field is measured. 2. **Identifying Units**: We need to identify the SI units of each term in the equation: - The SI unit of magnetic field \( B \) is Weber per square meter (Wb/m²). - The SI unit of current \( I \) is Ampere (A). - The unit of \( dl \) (length) is meter (m). - The sine function is dimensionless (unitless). - The distance \( r \) is also in meters (m). 3. **Rearranging the Equation**: Rearranging the formula to solve for \( \mu \): \[ \mu = B \cdot \frac{4\pi r^2}{I \cdot dl \cdot \sin \theta} \] 4. **Substituting Units**: Substitute the units into the rearranged equation: \[ \text{Unit of } \mu = \left( \frac{\text{Wb}}{\text{m}^2} \right) \cdot \frac{4\pi \cdot \text{m}^2}{\text{A} \cdot \text{m} \cdot \text{(unitless)}} \] 5. **Simplifying the Units**: Simplifying the expression: \[ \text{Unit of } \mu = \frac{\text{Wb} \cdot \text{m}^2}{\text{m}^2 \cdot \text{A}} = \frac{\text{Wb}}{\text{A} \cdot \text{m}} \] 6. **Final Answer**: Thus, the SI unit of magnetic permeability (μ) is: \[ \text{Wb} \cdot \text{A}^{-1} \cdot \text{m}^{-1} \] ### Summary: The SI unit of magnetic permeability is \( \text{Wb} \cdot \text{A}^{-1} \cdot \text{m}^{-1} \).

To determine the SI unit of magnetic permeability, we can start from the relationship between magnetic field (B), magnetic permeability (μ), current (I), and the geometry of the system. ### Step-by-Step Solution: 1. **Understanding the Formula**: We know that the magnetic field (B) can be expressed in terms of magnetic permeability (μ) and current (I). The formula is: \[ B = \frac{\mu}{4\pi} \cdot \frac{I \cdot dl \cdot \sin \theta}{r^2} ...
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DC PANDEY ENGLISH-MAGNETICS-Exercise
  1. A straight wilre of diameter 0.5 mm carrying a current 2A is replaced ...

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  2. The path of a charged particle moving in a uniform steady magnetic fie...

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  3. The SI unit of magnetic permebility is

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  4. Identify the correct statement bout the magnetic field lines.

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  5. Identify the correct sttement related to the direction of magnetic mom...

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  6. A non planar closed loop of arbitrary shape carryign a current I is pl...

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  7. The magnetic dipole moment of current loop is independent of

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  8. The accelertion of a electron at a certain moment in a magnetic field ...

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  9. A closed loop a current I lies in the xz-plane. The loop will experien...

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  10. A stream of protons and alpha-particle of equal momenta enter a unifom...

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  11. As loop of magnetic moment M is placed in the orientation of unstable ...

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  12. A current of 50 A is placed through a straight wire of length 6 cm th...

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  13. The magnetic field due to a current carrying circular loop of radius 3...

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  14. A conductor ab of arbitrary shape carries current I flowing from b to ...

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  15. When an electron is accelerated through a potential difference V, it e...

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  16. Two long straight wires, each carrying a current I in opposite directi...

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  17. The magnetic field ast a distance x on the axis of a circular coil of ...

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  18. Electric field and magnetic field n a region of space is given by E=E0...

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  19. An electron having kinetic energy K is moving i a circular orbit of ra...

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  20. Four long staight wires are located at the corners of a square ABCD. A...

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