Magnetism And Matter – JEE Main PhysicsPractice Questions & PYQs

From the given B-H curve, the coercivity, which is the magnetic field intensity $H$ required to reduce the magnetic induction $B$ to zero, is $H = 100 \text{ A/m}$.

The magnetic field intensity $H$ inside a long solenoid is given by $H = nI$, where $n$ is the number of turns per unit length.

Converting the turn density to SI units:
$n = 1000 \text{ turns/cm} = 1000 \times 100 \text{ turns/m} = 10^5 \text{ m}^{-1}$

Using the formula to solve for current $I$:
$100 \text{ A/m} = (10^5 \text{ m}^{-1}) I$
$I = \frac{100}{10^5} \text{ A} = 10^{-3} \text{ A} = 1 \text{ mA}$

$T=2\pi \sqrt{1/{MB}}$ $I=7.5\times 10^{-6}kg-m^2$ $M=6.7\times 10^{-2}Am^2$By substituting value in the formulaT=.665 secTime = 10 T = 6.65 sec

$V_B=B_4(5)(240)$ $B_H=B\text{cos}\theta$ $B_H={5\sqrt5\times 10^{-5}}/3$ $B_v=10/3\times 10^{-5}$T $V_B={5\sqrt5}/3\times 10^{-5}\times 5\times 240$ $V_B=44.6mV=45mV$ $v_w=B_vlV$ $=10^{-4}\times 1200$ $V_\omega =120$mV(left side at fighter voltage)

For electromagnet and transformer, the coercivity should be low to reduce energy loss

$\mu _0H=\mu _0ni$ $3\times 10^3=100/0.1\times i\Rightarrow =3\;A$

$B_{net}=B_{M_1}+B_{M_2}+B_H$ $\frac{\mu _0M_1}{4\pi x^3}+\frac{\mu _0M_2}{4\pi x^3}+B_H$ $=\mu _0/{4\pi x^3}(M_1+M_2)+B_H$ $=10^{-7}/10^{-3}\times 2.2+3.6\times 10^{-5}$ $=2.56\times 10^{-4} {Wb}/m^2$

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.

KEY CONCEPT : The temperature above which a ferromagnetic substance becomes paramagnetic is called Curie's temperature.

$T=2 \pi \sqrt{\;\frac{1}{M \times B}}$ where $I=\;\frac{1}{12} m ℓ^2$ When the magnet is cut into three pieces the pole strength will remain the same and M. I. $(I^{'}) =\;\frac{1}{12} (\;\frac{m}{3}) (\;\frac{ℓ}{3}) ^2 \times 3=\;\frac{I}{9}$ We have, Magnetic moment $(M)$ $=$ Pole strength $(m) \times ℓ$ $\therefore$ New magnetic moment, $=$ Pole strength $(m) \times ℓ$ $\therefore$ New magnetic moment,$M^{'}=m \times (\;\frac{ℓ}{3}) \times 3=m ℓ=M$ $\therefore T^{'}=\;{T}/{{\sqrt{9}}}=\;\frac{2}{3} s .$

NOTE : Electro magnet should be amenable to magnetization & demagnetization.$\therefore$ retentivity should be low $\&$ coercivity should be low

$W=M B (\cos \theta _1-\cos \theta _2)$ $=M B (\cos \theta ^{\circ}-\cos 60^{\circ})$ $=M B (1-\;\frac{1}{2}) =\;\frac{M B}{2}$ $\therefore \tau =M B sin \;\theta =M B sin \;60^{\circ}$ $={\sqrt{3}} \;\frac{M B}{2}={\sqrt{3}} W$

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

KEY CONCEPT : The time period of a rectangular magnet oscillating in earth's magnetic field is given by $T=2 \pi \sqrt{\;\frac{I}{\mu B_H}}$ where $I=$ Moment of inertia of the rectangular magnet$\mu =$ Magnetic moment$B_H=$ Horizontal component of the earth's magnetic field Case $1: T=2 \pi \sqrt{\;\frac{I}{\mu B_H}}$ where $I=\;\frac{1}{12} M ℓ^2$ Case 2 : Magnet is cut into two identical pieces such that each piece has half the original length. Then $T^{'}=2 \pi \sqrt{\;\frac{I^{'}}{\mu ^{'} B_H}}$ where $I^{'}=\;\frac{1}{12} (\;\frac{M}{2}) (\;\frac{ℓ}{2}) ^2=\;\frac{I}{8}$ and $\mu ^{'}=\;\frac{\mu }{2}$ $\; \therefore \;\frac{T^{'}}{T}=\sqrt{\;\frac{I^{'}}{\mu ^{'}} \times \;\frac{\mu }{I}}$ $\;=\sqrt{\;\frac{I / 8}{\mu / 2} \times \;\frac{\mu }{I}}$ $\;=\sqrt{\;\frac{1}{4}}=\;\frac{1}{2}$