1. A paramagnetic material has a magnetic field intensity of 10⁴ Am^-1. If the susceptibility of the material at room temperature is 3.7×10^-3. Calculate the magnetization and flux density in the material.

Given

H = 10⁴ A m^-1, χ = 3.7 × 10^-3, I =?, B = ?

Solution:-

(i) \begin{aligned} \chi & =\frac{\mathrm{I}}{\mathrm{H}} \\ \mathrm{I} & =\chi \mathrm{H} \\ & =3.7 \times 10^{-3} \times 10^{4} \\ & =3.7 \times 10 \\ \mathrm{I} & =37 \mathrm{Am}^{-1} \end{aligned}

(ii) \begin{aligned} \mathrm{B} & =\mu_{\mathrm{o}}(\mathrm{I}+\mathrm{H}) \\ & =4 \pi \times 10^{-7} \times\left(37+10^{4}\right) \\ & =126179.4 \times 10^{-7} \\ \mathrm{~B} & =0.0126 \mathrm{Wbm}^{-2} \end{aligned}

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2. A magnetic material has a magnetization of 2300 Am⁻¹ and produces a flux density of 0.00314 Wb m⁻². Calculate the magnetizing force and the relative permeability of the material.

Given

I = 23000 Am^-1, B = 0.00314Wb m^-2, H =?, μ_r = ?

Solution:-

\begin{aligned} \mathrm{B} & =\mu_{\mathrm{o}}(\mathrm{I}+\mathrm{H}) \\ \mathrm{H} & =\frac{\mathrm{B}}{\mu_{0}}-\mathrm{I} \\ & =\frac{0.00314}{4 \pi \times 10^{-7}}-2300 \\ \mathrm{H} & =198.7326 \mathrm{~A} \mathrm{~m}^{-1} \\ \chi & =\frac{\mathrm{I}}{\mathrm{H}}=\left(\mu_{\mathrm{r}}-1\right) \\ \mu_{\mathrm{r}} & =\frac{\mathrm{I}}{\mathrm{H}}+1 \\ & =\frac{2300}{198.7326}+1 \\ \mu_{\mathrm{r}} & =12.573 \end{aligned}

μ_r = 12.573

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3. A paramagnetic material has BCC structure with a cubic edge of 2.5 A°. If the saturation value of magnetization is 1.8×10⁶ Am⁻¹, Calculate the magnetization contributed per atom in Bohr magnetons.

Given

a = 2.5 × 10^-10 m, I = 1.8×10⁶ Am^-1, M_T = ?

Solution:-

The number of atoms present per unit volume

\begin{aligned} \mathrm{N} & =\frac{\text { Number of atoms present in an unit cell }}{\text { Volume of the unit cell }} \\ & =\frac{2}{\left(2.5 \times 10^{-10}\right)^{3}} \\ \mathrm{~N} & =1.28 \times 10^{29} \mathrm{~m}^{-3} \end{aligned}

The magnetization produced per atom =\frac{\text { Totalmagnetisation }}{\text { No. of atomsper unit volun }} \begin{aligned} & =\frac{\mathrm{I}}{\mathrm{N}} \\ & =\frac{1.8 \times 10^{6}}{1.28 \times 10^{29}} \\ \mathrm{I}_{\mathrm{T}} & =1.40625 \times 10^{-23} \mathrm{Am}^{-2} \\ \mu_{\mathrm{B}} & =\frac{\mathrm{eh}}{4 \pi \mathrm{m}} \\ & =\frac{1.6 \times 10^{-19} \times 6.625 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31}} \\ \beta & =9.27 \times 10^{-24} \mathrm{~A} \mathrm{~m}^{-2} \end{aligned}

∴ Magnetisation produced per atom in Bohr magneton

\begin{aligned} & =\frac{1.40625 \times 10^{-23}}{9.27 \times 10^{-24}} \\ & =1.516 \mathrm{Bohr} \text { magneton } \end{aligned}

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4. The saturation magnetic induction of Nickel is 0.65 Wb m⁻². If the density of Nickel is 8906 Kg m⁻³ and its atomic weight is 58.7. Calculate the magnetic moment of the Nickel atom in Bohr magneton.

Given

B_s = 0.65 Wb m^-2, ρ = 8906 Kg m^-3, Atomic Weight = 58.7, μ_m = ?

Solution:-

\begin{aligned} N & =\frac{\rho \times \text { Avogadra number }}{\text { Atomic weight }} \\ & =\frac{8906 \times 6.023 \times 10^{26}}{58.7} \\ N & =9.14 \times 10^{28} \text { atoms } / \mathrm{m}^{-3} \end{aligned}

When χ is very large, Saturation magnetization

\begin{aligned} B_s & =N \mu_0 \mu_m \\ \mu_m & =\frac{B_s}{N \mu_o} \\ & =\frac{0.65}{9.14 \times 10^{26} \times 4 \times 3.14 \times 10^{-7}} \\ \mu_{\mathrm{m}} & =5.66 \times 10^{-24} \mathrm{Am}^2 \end{aligned}

Magnetic moment per Bohr magneton

\therefore \mu_{\mathrm{m}}=\frac{5.66 \times 10^{-24}}{9.29 \times 10^{-24}} Bohr magneton

μ_m = 0.61 μ_B

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5. In a magnetic material the field strength is found to be 10⁶ Am⁻¹. If the magnetic susceptibility of the material is 0.5×10⁻⁵. Calculate the intensity of magnetization and flux density in the material.

Given

H = 10⁶ Am^-1, χ = 0.5 × 10^-5, B = ?, I = ?

Solution:-

(i) I = χH

= 0.5 × 10⁶ × 10^-5

I = 5 A m^-1

(ii) B = μ₀ (I + H)

= 4 × 3.14 × 10^-7 (5 + 10⁶)

B = 1.257 Wbm^-2

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6. The critical temperature of Niobium is 9.15 K. At 0K the critical field is 0.196 T. Calculate the critical field at 5 K.

Given

T_c = 9.15 K, T = 5 K, H₀ = 0.196 T, H_C = ?

Solution:-

\mathrm{H}_{\mathrm{c}}=\mathrm{H}_0\left(1-\frac{\mathrm{T}^2}{\mathrm{~T}_{\mathrm{c}}^2}\right) \begin{aligned} & =0.196\left[1-\frac{5^2}{9.15^2}\right] \\ & =0.196 \times 0.7014 \\ c_c & =\mathrm{OH} 374 \text { Tesla } \end{aligned}

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7. Calculate the critical current through a long thin superconducting wire of radius 0.5 mm. The critical magnetic field is 42.75×10³ A/m.

Given

r = 0.5 × 10^-3 m, H_c = 42.75 × 10³ A/m, I_c = ?

Solution:-

I_c = 2πr H_c

= 2×3.14×0.5×10^-3 × 42.75 × 10³

I_c = 134.25 Ampere

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8. Calculate the critical current for a superconducting wire of lead having a diameter of 1 mm at 4.2K. Critical temperature for lead is 7.18 K and H_c(0) = 6.5 × 10⁴ A/m.

Given

r = 0.5 × 10^-3 m, T = 4.2 K, T_c = 7.18 K, H₀ = 6.5 × 10⁴ A/m, I_c = ?, H_c = ?

Solution:-

\begin{aligned} \mathrm{H}_c & =\mathrm{H}_{\mathrm{o}}\left(1-\frac{\mathrm{T}^2}{\mathrm{~T}_{\mathrm{C}}^2}\right) \\ & =6.5 \times 10^4\left(1-\frac{4.2^2}{7.18^2}\right) \\ & =6.5 \times 10^4(1-0.3421) \\ \mathrm{H}_{\mathrm{c}} & =42.758 \times 10^3 \text { Tesla } \\ \mathrm{I}_{\mathrm{c}} & =2 \pi \mathrm{r} \mathrm{H}_{\mathrm{c}} \\ & =2 \times 3.14 \times 0.5 \times 10^{-3} \times 42.758 \times 10^3 \\ \mathrm{I}_{\mathrm{c}} & =134.26 \text { Ampere } \end{aligned}

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9. A voltage of 5.9×10⁻⁶ V is applied across a Josephson junction. What is the frequency of the radiation emitted by the junction?

Given

V = 5.9 × 10^-6 V, v=?

Solution:-

\begin{aligned} \mathrm{v} & =\frac{2 \mathrm{eV}}{\mathrm{h}} \\ & =\frac{2 \times 1.6 \times 10^{-19} \times 5.9 \times 10^{-6}}{6.62 \times 10^{-34}} \\ v & =2.851 \times 10^9 \mathrm{~Hz} \end{aligned}

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10. Prove that susceptibility of superconductor is -1 and relative permeability is zero.

Solution:-

eqn. 1

We know, the induced magnetic field B = μ_o (I+H)

In superconductor, B = 0.

Therefore eqn. (1) can be written as

0 = μ_o (I+H)      eqn. 2

Since, μ_o #0,

I = -H

We know I/H = χ = -1 eqn. 3

Also,  χ = μ_r = -1 eqn. 4

Equating eqn. (3) and (4) we get

μ_r -1 = -1

μ_r = 1  -1

μ_r = 0

Substituting eqn. (5) in (4) we get