Magnetic Properties of Matter

A perfect diamagnetic substance will completely expel the magnetic field. Therefore, there will be no magnetic field inside the cavity of sphere. Hence the paramagnetic substance kept inside the cavity will experience no force.

Option (2) is correct as permanent magnet should have high coercivity & retentivity.

Retentivity = 1.0 T

Co-ercivity = 50 A/m

Saturation = 1.5 T

${f_1} = {1 \over {2\pi }}\sqrt {{{\mu {B_1}\cos {{45}^o}} \over I}} $

${f_2} = {1 \over {2\pi }}\sqrt {{{\mu {B_2}\cos {{30}^o}} \over I}} $

${{{f_1}} \over {{f_2}}} = {{{B_1}\cos {{45}^o}} \over {{B_1}\cos {{30}^o}}}$

$ \therefore $ ${{{B_1}} \over {{B_2}}} = 0.7$

${\overrightarrow F _m} = q\left( {\overrightarrow V \times \overrightarrow B } \right)$

$\overrightarrow B = {\overrightarrow B _x} + {\overrightarrow B _y}$ Since M_y = 2M_x

$ \Rightarrow \left| {{{\overrightarrow B }_x}} \right| = \left| {{{\overrightarrow B }_y}} \right|$

${\overrightarrow B _{net}}$ is parallel to $\overrightarrow V $

$ \Rightarrow $ $\overrightarrow F = 0$

x $\alpha $ ${1 \over {{T_C}}}$

curie law for paramagnetic substane

${{{x_1}} \over {{x_2}}}$ = ${{{T_{{C_2}}}} \over {{T_{{C_1}}}}}$

${{2.8 \times {{10}^{ - 4}}} \over {{x_2}}} = {{300} \over {350}}$

x₂ = ${{2.8 \times 350 \times {{10}^{ - 4}}} \over {300}}$

= 3.266 $ \times $ 10^$-$4

x = ${1 \over H}$

I = ${{Magnetic\,moment} \over {Volume}}$

I = ${{20 \times {{10}^{ - 6}}} \over {{{10}^{ - 6}}}}$ = 20 N/m²

x = ${{20} \over {60 \times {{10}^{ + 3}}}}$ = ${1 \over 3} \times {10^{ - 3}}$

= 0.33 $ \times $ 10^$-$3 = 3.3 $ \times $ 10^$-$4

Without applied forces, (in equilibrium position) the needle will stay in the resultant magnetic field of earth. Hence, the dip ' $\theta$ ' at this place is $45^{\circ}$ (given).


We know that, horizontal and vertical components of earth's magnetic field $\left(B_H\right.$ and $B_V$ ) are related as

$ \frac{B_V}{B_H}=\tan \theta $

Here, $\theta=45^{\circ}$ and $B_H=18 \times 10^{-6} \mathrm{~T}$

$ \Rightarrow B_V=B_H \tan 45^{\circ} $

$ \Rightarrow B_V=B_H=18 \times 10^{-6} \mathrm{~T} \quad\left(\because \tan 45^{\circ}=1\right)$

Now, when the external force $F$ is applied, so as to keep the needle stays in horizontal position is shown below,


Taking torque at point $P$, we get

$ \begin{array}{ll} m B_V \times 2 l=F l \\\\ \therefore F=2 \times m B_V \end{array} $

Substituting the given values, we get

$ \begin{aligned} & =2 \times 1.8 \times 18 \times 10^{-6} \\\\ & =6.48 \times 10^{-5}=6.5 \times 10^{-5} \mathrm{~N} \end{aligned} $

To determine the work done in reversing the direction of a magnetic moment in a time-varying magnetic field, we'll follow these steps:

Given:

Magnetic moment: $\mathbf{m} = 10^{-2} \hat{i}$ A·m²

Magnetic field: $\mathbf{B}(t) = B \cos(\omega t) \hat{i}$, where $B = 1$ T and $\omega = 0.125$ rad/s

Time at which reversal occurs: $t = 1$ s

Step 1: Calculate the Magnetic Field at $t = 1$ s

$ B(t) = B \cos(\omega t) = 1 \times \cos(0.125 \times 1) = \cos(0.125 \text{ rad}) $

Compute $\cos(0.125 \text{ rad})$:

$ \cos(0.125 \text{ rad}) \approx 0.9922 $

So,

$ B(t) \approx 0.9922 \text{ T} $

Step 2: Calculate the Work Done

The potential energy $U$ of a magnetic dipole in a magnetic field is:

$ U = -\mathbf{m} \cdot \mathbf{B} = -mB \cos\theta $

The work done $W$ in reversing the magnetic moment from $\theta = 0^\circ$ to $\theta = 180^\circ$ is:

$ W = U_{\text{final}} - U_{\text{initial}} = [-mB \cos(180^\circ)] - [-mB \cos(0^\circ)] = mB (\cos 0^\circ - \cos 180^\circ) $

Simplify:

$ W = mB (1 - (-1)) = 2mB $

Substitute the values:

$ W = 2 \times (10^{-2} \text{ A·m}^2) \times (0.9922 \text{ T}) \approx 0.01984 \text{ J} $

Step 3: Approximate Using RMS Value

Given that the magnetic field is time-varying, we can consider the root mean square (RMS) value of $\cos(\omega t)$ over a complete cycle:

$ \text{RMS of } \cos(\omega t) = \frac{1}{\sqrt{2}} $

Thus, the RMS value of the magnetic field is:

$ B_{\text{RMS}} = B \times \frac{1}{\sqrt{2}} = 1 \times \frac{1}{\sqrt{2}} \approx 0.7071 \text{ T} $

Now, calculate the work done using $B_{\text{RMS}}$:

$ W_{\text{RMS}} = 2mB_{\text{RMS}} = 2 \times (10^{-2}) \times 0.7071 \approx 0.01414 \text{ J} $

Coercivity, H = ${B \over {{\mu _0}}}$

Inside solenoid the magnetic field,

B = $\mu $₀ni

$ \therefore $   H = ${{{\mu _0}ni} \over {{\mu _0}}}$

= ni

= ${N \over \ell } \times i$

= ${{100} \over {0.2}} \times 5.2$

= 2600 A/m

Given that,

In the solenoid there is 1000 turns in 1 cm.

$\therefore\,\,\,\,$ In 100 cm or 1 m no of turns

= 1000 $ \times $ 100 = 10⁵ turns/m

$\therefore\,\,\,\,$ No of turns, n = 10⁵

Correctivity of ferromagnet, H = 100 A/m

As we know,

H = n I

$\therefore\,\,\,\,$ I = ${{100} \over {{{10}^5}}} = 1mA$

$\therefore\,\,\,\,$ Current to demagnetise the ferromagnet = 1 mA.

Given : Magnetic moment, M = 6.7 × 10^–2 Am²

Magnetic field, B = 0.01 T

Moment of inertia, I = 7.5 × 10^–6 Kgm²

Using, T = $2\pi \sqrt {{I \over {MB}}} $

= $2\pi \sqrt {{{7.5 \times {{10}^{ - 6}}} \over {6.7 \times {{10}^{ - 2}} \times 0.01}}} $

= ${{2\pi } \over {10}} \times 1.06$ s

Time taken for 10 complete oscillations

t = 10T = 2$\pi $ × 1.06

= 6.6568 $ \simeq $ 6.65 s

V_B = vhBcos$\theta $

= 240 $ \times $ 5 $ \times $ 5 $ \times $ 10^$-$5$ \times $ ${{\sqrt 5 } \over 3}$

= 44.7 $ \times $ 10^$-$3 V

= 45 mV



V_w = $l$vB sin$\theta $

= 15 $ \times $ 240 $ \times $ 5 $ \times $ 10^$-$5 $ \times $ ${2 \over 3}$

= 1200 $ \times $ 10^$-$4 V

= 120 mV

From right hand rule, the charge moves to the left side of the pilot.

Graph $\left[ A \right]$ is for material used for making permanent magnets (high coercivity)

Graph $\left[ B \right]$ is for making electromagnets and transformers.

Key Concept:

• In a superconductor below its critical temperature, the Meissner effect occurs → it completely expels magnetic flux from its interior.

• This means the magnetic induction inside the superconductor is:

$ B_s = 0 $

regardless of the applied external field (until the critical field strength is exceeded).

Correct Option:

Option B: $ B_s = 0 $ ✅

Magnetic field in solenoid $B = {\mu _0}ni$

$ \Rightarrow {B \over {{\mu _0}}} = ni$

(Where $n=$ number of turns per unit length)

$ \Rightarrow {B \over {{\mu _0}}} = {{Ni} \over L}$

$ \Rightarrow 3 \times {10^3} = {{100i} \over {10 \times {{10}^{ - 2}}}}$

$ \Rightarrow i = 3A$

Given: ${M_1} = 1.20A{m^2}\,\,\,$ and $\,\,\,{M_2} = 1.00A{m^2}$

$r = {{20} \over 2}cm = 0.1m$

${B_{net}} = {B_1} + {B_2} + {B_H}$

${B_{net}} = {{{\mu _0}\left( {{M_1} + {M_2}} \right)} \over {{r^3}}} + {B_H}$

$ = {{{{10}^{ - 7}}\left( {1.2 + 1} \right)} \over {{{\left( {0.1} \right)}^3}}} + 3.6 \times {10^{ - 5}}$

$ = 2.56 \times {10^{ - 4}}\,\,wb/{m^2}$

For a diamagnetic material, the value of ${\mu _r}$ is less than one. For any material, the value of ${ \in _r}$ is always greater than $1.$

Ferromagnetic substance has magnetic domains whereas para-magnetic substances have magnetic dipoles which get attracted to a magnetic field. Diamagnetic substances do not have magnetic dipole but in the presence of external magnetic field due to their orbital motion of electrons these substances are repelled.

A magnetic needle kept in non uniform magnetic field experience a force and torque due to unequal forces acting on poles.

$T = 2\pi \sqrt {{1 \over {M \times B}}} $ where $I = {1 \over {12}}m{\ell ^2}$

When the magnet is cut into three pieces the pole strength will remain the same and

$M.{\rm I}.\left( {I'} \right) = {1 \over {12}}\left( {{m \over 3}} \right){\left( {{\ell \over 3}} \right)^2} \times 3 = {I \over 9}$

We have, Magnetic moment $(M)$

$=$ Pole strength $\left( m \right) \times \ell $

$\therefore$ New magnetic moment,

$M' = m \times \left( {{\ell \over 3}} \right) \times 3 = m\ell = M$

$\therefore$ $T' = {T \over {\sqrt 9 }} = {2 \over 3}s.$

The Curie temperature is the temperature above which a ferromagnetic material becomes paramagnetic.

Ferromagnetic materials have a high degree of magnetization in the presence of a magnetic field. Above the Curie temperature, these materials lose their ferromagnetic behavior and become paramagnetic, meaning they are weakly attracted to a magnetic field and do not retain any magnetization in the absence of an external magnetic field.

So, the correct option is :

Option A : a ferromagnetic material becomes paramagnetic.

Materials suitable for making electromagnets should have low retentivity and low coercivity.

Retentivity (or remanence) is the ability of a magnetic material to retain its magnetism after the removal of the magnetizing force. For an electromagnet, we want this to be low, as we want the magnet to only be magnetic when current is flowing.

Coercivity is the ability of a magnetic material to resist becoming demagnetized. Again, for an electromagnet, we want this to be low, as we want to easily turn off the magnetism when the current is removed.

So, the correct option is :

Option B: Low retentivity and low coercivity.

KEY CONCEPT : The time period of a rectangular magnet oscillating in earth's magnetic field is given by

$T = 2\pi \sqrt {{I \over {\mu {B_H}}}} $

where $I=$ Moment of inertia of the rectangular magnet

$\mu = $ Magnetic moment

${B_H} = $ Horizontal component of the earth's magnetic field

Case 1 : $T = 2\pi \sqrt {{I \over {\mu {B_H}}}} $ where $I = {1 \over {12}}M{\ell ^2}$

Case 2 : Magnet is cut into two identical pieces such that each piece has half the original length. Then

$T' = 2\pi \sqrt {{{I'} \over {\mu '{B_H}}}} $

where $I' = {1 \over {12}}\left( {{M \over 2}} \right){\left( {{\ell \over 2}} \right)^2} = {I \over 8}$ and $\mu ' = {\mu \over 2}$

$\therefore$ ${{T'} \over T} = \sqrt {{{I'} \over {\mu '}} \times {\mu \over I}} $

$ = \sqrt {{{I/8} \over {\mu /2}} \times {\mu \over I}} $

$ = \sqrt {{1 \over 4}} = {1 \over 2}$

$W = MB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$

$ = MB\left( {\cos {\theta ^ \circ } - \cos {{60}^ \circ }} \right)$

$ = MB\left( {1 - {1 \over 2}} \right) = {{MB} \over 2}$

$\therefore$ $\tau = MB\,\sin \theta = MB\,\sin \,{60^ \circ }$

$ = \sqrt 3 {{MB} \over 2} = \sqrt 3 W$

As shown in the figure, the magnetic lines of force are directed from south to north inside a bar magnet.