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Magnetism and Matter - Magnetisation and Magnetic Intensity

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Magnetisation (MM): It is defined as the net magnetic moment per unit volume of a material. If mnetm_{net} is the net magnetic moment developed in a volume VV, then M=mnetVM = \frac{m_{net}}{V}. Its SI unit is A m−1A \, m^{-1}.

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Magnetic Intensity (HH): It represents the degree to which an external magnetic field can magnetise a material. For a solenoid with nn turns per unit length carrying current II, the magnetic intensity is H=nIH = nI. Its SI unit is A m−1A \, m^{-1}.

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Magnetic Susceptibility (χ\chi): It is a measure of how easily a substance can be magnetised when placed in a magnetising field. It is the ratio of magnetisation to magnetic intensity: χ=MH\chi = \frac{M}{H}. It is a dimensionless quantity.

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Relation between B,H,B, H, and MM: The total magnetic field BB inside a material is the sum of the magnetic field due to external current and the field due to the magnetisation of the material: B=μ0(H+M)B = \mu_0 (H + M).

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Magnetic Permeability (μ\mu): It is the ability of a material to allow magnetic lines of force to pass through it. The relative permeability is μr=μμ0=1+χ\mu_r = \frac{\mu}{\mu_0} = 1 + \chi.

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Curie's Law: For paramagnetic materials, the magnetic susceptibility is inversely proportional to the absolute temperature TT: χ=CT\chi = \frac{C}{T}, where CC is Curie's constant.

📐Formulae

M=mnetVM = \frac{m_{net}}{V}

H=nIH = nI

B=μ0(H+M)B = \mu_0 (H + M)

M=χHM = \chi H

μr=1+χ\mu_r = 1 + \chi

B=μ0μrHB = \mu_0 \mu_r H

χ=CT\chi = \frac{C}{T}

💡Examples

Problem 1:

A solenoid has a core of material with relative permeability μr=400\mu_r = 400. The windings of the solenoid are insulated from the core and carry a current such that the magnetic intensity H=2000 A m−1H = 2000 \, A \, m^{-1}. Calculate the magnetisation MM of the core.

Solution:

Given: μr=400\mu_r = 400 and H=2000 A m−1H = 2000 \, A \, m^{-1}. We know that μr=1+χ\mu_r = 1 + \chi, so the susceptibility χ\chi is: χ=μr−1=400−1=399\chi = \mu_r - 1 = 400 - 1 = 399 Now, the magnetisation is given by M=χHM = \chi H: M=399×2000M = 399 \times 2000 M=798000 A m−1M = 798000 \, A \, m^{-1}

Explanation:

The magnetisation is found by multiplying the magnetic susceptibility of the core by the applied magnetic intensity. Since μr\mu_r is very high, the material is likely ferromagnetic.

Problem 2:

A sample of paramagnetic salt has a magnetic susceptibility of 2.4×10−42.4 \times 10^{-4} at a temperature of 300 K300 \, K. Calculate its susceptibility at 200 K200 \, K using Curie's Law.

Solution:

According to Curie's Law, χ∝1T\chi \propto \frac{1}{T}, or χ1T1=χ2T2\chi_1 T_1 = \chi_2 T_2. Given: χ1=2.4×10−4\chi_1 = 2.4 \times 10^{-4}, T1=300 KT_1 = 300 \, K, T2=200 KT_2 = 200 \, K. χ2=χ1T1T2\chi_2 = \frac{\chi_1 T_1}{T_2} χ2=2.4×10−4×300200\chi_2 = \frac{2.4 \times 10^{-4} \times 300}{200} χ2=3.6×10−4\chi_2 = 3.6 \times 10^{-4}

Explanation:

As temperature decreases, the thermal agitation of molecular dipoles decreases, allowing them to align more easily with the external field, thus increasing susceptibility.

Problem 3:

Calculate the difference in magnetic field BB (in Tesla) when the magnetisation MM increases from 500000500000 to 800000800000 A/mA/m in a vacuum (assume HH is constant and μ0≈4π×10−7 T m/A\mu_0 \approx 4\pi \times 10^{-7} \, T \, m/A). For simplification of calculation, find the difference in MM first.

Solution:

To find the difference in MM: 800000−500000300000\begin{array}{r} 800000 \\ -500000 \\ \hline 300000 \end{array} Difference in M=300000 A m−1M = 300000 \, A \, m^{-1}. The change in BB is ΔB=μ0ΔM\Delta B = \mu_0 \Delta M. ΔB=4π×10−7×3×105\Delta B = 4\pi \times 10^{-7} \times 3 \times 10^5 ΔB=1.2π×10−1≈0.377 T\Delta B = 1.2\pi \times 10^{-1} \approx 0.377 \, T

Explanation:

The change in total magnetic field is directly proportional to the change in magnetisation when the external intensity HH remains constant.