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An iron rod of volume 10^(-4)m^(3) and r...

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

A

`20Am^(2)`

B

`25Am^(2)`

C

`30Am^(2)`

D

`35Am^(2)`

Text Solution

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The correct Answer is:
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: Understand the parameters given - Volume of the iron rod, \( V = 10^{-4} \, m^3 \) - Relative permeability of the iron rod, \( \mu_r = 1000 \) - Turns per unit length of the solenoid, \( n = 5 \, \text{turns/cm} = 500 \, \text{turns/m} \) - Current through the solenoid, \( I = 0.5 \, A \) ### Step 2: Calculate the magnetic field inside the solenoid The magnetic field \( B \) inside a solenoid is given by the formula: \[ B = \mu_0 (H + I) \] Where: - \( \mu_0 \) is the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \) - \( H \) is the magnetic field strength, given by \( H = nI \) First, calculate \( H \): \[ H = n \cdot I = 500 \, \text{turns/m} \cdot 0.5 \, A = 250 \, A/m \] Now, substitute \( H \) into the equation for \( B \): \[ B = \mu_0 H = (4\pi \times 10^{-7}) \cdot 250 \] Calculating \( B \): \[ B = 4\pi \times 10^{-7} \cdot 250 \approx 3.14 \times 10^{-4} \, T \] ### Step 3: Calculate the magnetic permeability of the iron rod The magnetic permeability \( \mu \) of the iron rod can be calculated using: \[ \mu = \mu_r \cdot \mu_0 \] Substituting the values: \[ \mu = 1000 \cdot (4\pi \times 10^{-7}) \approx 4\pi \times 10^{-4} \, T \cdot m/A \] ### Step 4: Calculate the intensity of magnetization \( I \) The intensity of magnetization \( I \) can be calculated using: \[ I = \frac{\mu - \mu_0}{\mu_0} \cdot H \] Substituting the values: \[ I = \left(\frac{4\pi \times 10^{-4} - 4\pi \times 10^{-7}}{4\pi \times 10^{-7}}\right) \cdot 250 \] Calculating \( I \): \[ I \approx (1000 - 1) \cdot 250 \approx 999 \cdot 250 \approx 249750 \, A/m \] ### Step 5: Calculate the magnetic moment \( M \) The magnetic moment \( M \) is given by: \[ M = I \cdot V \] Substituting the values: \[ M = 249750 \cdot 10^{-4} \approx 24.975 \, A \cdot m^2 \] ### Final Answer Rounding to two decimal places, the magnetic moment of the rod is approximately: \[ M \approx 25 \, A \cdot m^2 \]

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: Understand the parameters given - Volume of the iron rod, \( V = 10^{-4} \, m^3 \) - Relative permeability of the iron rod, \( \mu_r = 1000 \) - Turns per unit length of the solenoid, \( n = 5 \, \text{turns/cm} = 500 \, \text{turns/m} \) - Current through the solenoid, \( I = 0.5 \, A \) ...
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