PhysicsKund Physics class 9 to 12 NCERT, JEE - NEET

Notes : CBSE Class 12 Physics Chapter 5 Magnetism and Matter - Physics Kund 1. Bar Magnet 2. Bar Magnet as an Equivalent Solenoid 3. Magnetic Field Intensity Due to a Magnetic Dipole (Bar Magnet) Along its Axis 4. Magnetic Field Intensity Due to a Magnetic Dipole (Bar Magnet) Along the Perpendicular (Equatorial) Line 5. Torque on a Magnetic Dipole (Bar Magnet) in a Uniform Magnetic Field 6. Magnetic Field Lines 7. Magnetic Properties of Materials – Para-, Dia- and Ferro-magnetic Substances with Examples 8. Magnetization of Materials 9. Effect of Temperature on Magnetic Properties

Define magnetic flux. State its S.I. unit and write its dimensional formula  Definition of Magnetic Flux : It is defined as the number of magnetic field lines passing through a surface. It is denoted by . Consider a surface S in a uniform magnetic field $\vec{B}$. Let the surface is made up of small elements each of area vector $\vec{dS}$ . If the magnetic field makes an angle with the area vector $\vec{dS}=(ds) \hat{n}$ then the magnetic flux linked with the small element is given by $d\phi_B = \vec{B} \cdot d\vec{S} \quad ...(1)$ The magnetic flux through the surface S is equal to the sum of magnetic flux linked with all elements each of area vector $\vec{dS}$ of the surface. That is, $\phi_B = \sum \vec{B} \cdot d\vec{S} \quad ...(2)$ If the elements are very small in area and extremely large in number, then the magnetic flux linked with the closed surface is written as $\phi_B = \oint_S \vec{B} \cdot d\vec{S} \quad ...(3)$ S.I Unit of Magnetic Flux :  S.I. unit of magne...

Derivation : Potential energy of a magnetic dipole placed in magnetic field. Potential energy of magnetic dipole:   Work done to rotate a magnetic dipole in a uniform magnetic field is stored as potential energy of the magnetic dipole. If a magnetic dipole of magnetic dipole moment \( \vec{m} \) is placed at an angle \( \theta \) with respect to uniform magnetic field of strength \( \vec{B} \), then torque experienced by it is given by: \[\tau = mB \sin \theta\] If the dipole rotates through an angle \( d\theta \), then work done on it is: \[dW = \tau d\theta = mB \sin \theta d\theta\] Total work done to rotate the dipole from \( \theta_1 \) to \( \theta_2 \) position is: $W = \int_{\theta_1}^{\theta_2} mB \sin \theta d\theta$ $W= mB \int_{\theta_1}^{\theta_2} \sin \theta d\theta$ $W= mB [-\cos \theta]_{\theta_1}^{\theta_2}$ $W= -mB [\cos \theta_2 - \cos \theta_1]$ This work done is stored as the potential energy \( U \) of the magnetic dipole: $\therefore U = W = -mB (\cos \...

MAGNETIC DIPOLE (BAR MAGNET) IN A UNIFORM MAGNETIC FIELD :  Derive an expression for torque acting on a magnetic dipole placed in magnetic field. Consider a magnetic dipole (a bar magnet) placed in a uniform magnetic field \( \vec{B} \) such that the angle between the direction of magnetic dipole moment \( (\vec{m}) \) and the direction of magnetic field \( (\vec{B}) \) is \( \theta \)  Let:   - Magnetic Length of magnet = \( 2l \)   - Pole Strength of each pole = \( q_m \)   - Force acting on North pole = \( q_m B \) along the direction of \( \vec{B} \)   - Force acting on South pole = \( q_m B \) opposite to the direction of \( \vec{B} \) These two equal and opposite forces constitute a couple which tends to rotate the magnet in the direction of \( \vec{B} \). Thus, the bar magnet experiences a torque and tends to rotate. However, net force acting on the magnetic dipole is zero and hence magnetic dipole does not have the translational ...

Magnetic Field Intensity at a Point on Equatorial Line of Bar Magnet : Let (P) be a point on the equatorial line of a bar magnet such that its distance from centre (O) is (r). Magnetic field intensity at ( P ) due to N-pole of the bar magnet: $\vec{B_1} = \frac{\mu_0}{4\pi} \cdot \frac{q_m}{(\sqrt{r^2 + l^2})^2} \quad \text{along PC}$ $= \frac{\mu_0}{4\pi} \cdot \frac{qm}{(r^2 + l^2)} \quad \text{along PC} \quad ...(i)$ Magnetic field intensity at ( P ) due to S-pole of the bar magnet: $\vec{B_2} = \frac{\mu_0}{4\pi} \cdot \frac{q_m}{(r^2 + l^2)} \quad \text{along PS} \quad ...(ii)$ Resultant Magnetic Field on Equatorial Line $ ( \vec{B_1}$ ) and $( \vec{B_2} )$ are inclined at an angle $(2\theta)$.   Therefore, the resultant of these two field intensities is: $B_e = \sqrt{B1^2 + B2^2 + 2B1B_2 \cos 2\theta}$ Since $( |\vec{B1}| = |\vec{B2}|$: $B_e = \sqrt{2B1^2 + 2B_1^2 \cos 2\theta}$ $= \sqrt{2B_1^2 (1 + \cos 2\theta)}$ $(\because 1 + \cos 2\theta = 2 \cos^2 \theta)$ $= \sqrt...

Question : Derivation : Magnetic Field Intensity at a Point on the Axial Line of a Bar Magnet - Magnetism and Matter  Proof :  Let O be the centre of a bar magnet having magnetic length 2l. Let p be the point on the axial line of the bar magnet at a distance r from the centre O of the bar magnet. Let $q_{m}$ be the pole strength of each pole of the magnet. Let a unit north pole be placed at point P.  Magnetic field intensity at P due to north pole of the bar magnet, $\overrightarrow{B_{1}}$ = Force experienced by unit north pole at P due to north pole of the magnet , $$\overrightarrow{B_{1}} = \frac{F}{q_{0}}= \frac{\frac{\mu_{0}}{4\pi}\frac{q_{m}.q_{0}}{(r-l)^{2}}}{q_{0}}$$ $$\overrightarrow{B_{1}} = \frac{\mu_{0}}{4 \pi}\frac{q_{m}}{(r-l)^{2}} along NP$$ $$\overrightarrow{B_{1}} = \frac{\mu_{0}}{4 \pi}\frac{q_{m}}{(r-l)^{2}}\hat{i}$$ similarly , magnetic field intensity at a point P due to south pole of the  bar magnet , $$\overrightarrow{B_{2}} = \frac{\mu_{0}}{4 ...