Question

The magnetizing field which produces a magnetic flux of \(22 \times 10^{-6} \, \text{Wb}\) in a metal bar of area of cross-section \(2 \times 10^{-5} \, m^2\) is (susceptibility of the metal = 699):

• A. 2500 A/m
• B. 1250 A/m
• C. 3750 A/m
• D. 5000 A/m

Hint:

Calculating the magnetic field strength requires understanding how material properties like susceptibility affect magnetic permeability.

Answer 1

The Correct Option is B

Using the magnetic flux \( \Phi \) and the cross-sectional area \( A \), we first calculate the flux density \( B \): \[ B = \frac{\Phi}{A} = \frac{22 \times 10^{-6} \, \text{Wb}}{2 \times 10^{-5} \, \text{m}^2} = 1.1 \, \text{T} \] Given the susceptibility \( \chi = 699 \), we find the relative permeability \( \mu_r \): \[ \mu_r = 1 + \chi = 700 \] The absolute permeability \( \mu \) is then: \[ \mu = \mu_0 \mu_r = (4\pi \times 10^{-7}) \times 700 = 0.879 \, \text{H/m} \] Now, solve for \( H \): \[ H = \frac{B}{\mu} = \frac{1.1}{0.879} \approx 1250 \, \text{A/m} \]