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An iron rod of volume 10^-4m^3 and relat...

An iron rod of volume `10^-4m^3` and relative permeability `1000` is placed inside a long solenoid wound with 5 turns/cm. If a current of `0*5A` is passed through the solenoid, find the magnetic moment of the rod.

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To solve the problem, we need to find the magnetic moment of an iron rod placed inside a solenoid. Here are the steps to arrive at the solution: ### Step 1: Identify Given Data - Volume of the iron rod, \( V = 10^{-4} \, \text{m}^3 \) - Relative permeability, \( \mu_r = 1000 \) - Turns per centimeter of the solenoid, \( n = 5 \, \text{turns/cm} = 500 \, \text{turns/m} \) - Current through the solenoid, \( I = 0.5 \, \text{A} \) ### Step 2: Calculate the Magnetizing Field (H) The magnetizing field \( H \) in a solenoid is given by: \[ H = n \cdot I \] Substituting the values: \[ H = 500 \, \text{turns/m} \times 0.5 \, \text{A} = 250 \, \text{A/m} \] ### Step 3: Calculate the Magnetic Field (B) The magnetic field \( B \) in the solenoid can be expressed as: \[ B = \mu_0 \cdot H \cdot \mu_r \] Where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \). Thus: \[ B = (4\pi \times 10^{-7}) \cdot (250) \cdot (1000) \] Calculating this: \[ B = (4\pi \times 10^{-7}) \cdot 250000 = 3.14 \times 10^{-1} \, \text{T} \approx 0.314 \, \text{T} \] ### Step 4: Calculate the Intensity of Magnetization (I) The intensity of magnetization \( I \) can be calculated using: \[ I = \frac{B}{\mu_0} - H \] Substituting the values: \[ I = \frac{0.314}{4\pi \times 10^{-7}} - 250 \] Calculating \( \frac{0.314}{4\pi \times 10^{-7}} \): \[ \frac{0.314}{4\pi \times 10^{-7}} \approx 2.5 \times 10^{5} \, \text{A/m} \] Thus: \[ I = 2.5 \times 10^{5} - 250 \approx 2.5 \times 10^{5} \, \text{A/m} \] ### Step 5: Calculate the Magnetic Moment (M) The magnetic moment \( M \) is given by: \[ M = I \cdot V \] Substituting the values: \[ M = (2.5 \times 10^{5}) \cdot (10^{-4}) = 25 \, \text{A m}^2 \] ### Final Answer The magnetic moment of the iron rod is \( M = 25 \, \text{A m}^2 \). ---

To solve the problem, we need to find the magnetic moment of an iron rod placed inside a solenoid. Here are the steps to arrive at the solution: ### Step 1: Identify Given Data - Volume of the iron rod, \( V = 10^{-4} \, \text{m}^3 \) - Relative permeability, \( \mu_r = 1000 \) - Turns per centimeter of the solenoid, \( n = 5 \, \text{turns/cm} = 500 \, \text{turns/m} \) - Current through the solenoid, \( I = 0.5 \, \text{A} \) ...
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Knowledge Check

  • An iron rod of volume 10^(-4)m^(3) and relative permeability 1000 is placed inside a long solenoid wound with 5 "turns"//"cm" . If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod is

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    B
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    C
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    D
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  • An iron rod of volume 10^(-4) m^(3) and relative permeability 900 is placed inside a long solenoid wound with 6 turns/cm. If a current of 0.4 A is passed through the solenoid, the magnetic moment of the rod is

    A
    0.216 `Am^(2)`
    B
    2.16 `Am^(2)`
    C
    21.6 `Am^(2)`
    D
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  • A solenoid has length 0.4cm , radius 1 cm and 400 turns of wire. If a current fo 5 A is passed through this solenoid, then what is the magnetic field inside the solenoid?

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    B
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    C
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