MHT CET · Physics · Magnetic Properties of Matter
The correct relation between total magnetic field (B), magnetic intensity (H), permeability of free space \(\left(\mu_0\right)\) and susceptibility \((\chi)\) is
- A \(\frac{B}{\mathrm{H}}=\mu_0(1-\chi)\)
- B \(\frac{\mathrm{B}}{\mathrm{H}}=\mu_0(1+\chi)^2\)
- C \(\frac{\mathrm{B}}{\mathrm{H}}=\mu_0(1+\chi)\)
- D \(\frac{\mathrm{B}}{\mathrm{H}}=\mu_0(1-\chi)^2\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{B}}{\mathrm{H}}=\mu_0(1+\chi)\)
Step-by-step Solution
Detailed explanation
The general relationship between \(\mathbf{B}, \mathrm{H}\), and the material properties is given by:
\(B=\mu_0(1+\chi) H\)
Where:
- \(\mathbf{B}\) is the magnetic flux density,
- H is the magnetic field intensity,
- \(\mu_0\) is the permeability of free space,
- X is the magnetic susceptibility.
\(B=\mu_0(1+\chi) H\)
Where:
- \(\mathbf{B}\) is the magnetic flux density,
- H is the magnetic field intensity,
- \(\mu_0\) is the permeability of free space,
- X is the magnetic susceptibility.
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