Param Himalaya - परम हिमालय

NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter - Param Himalaya 5.1 A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5x 10 J. What is the magnitude of magnetic moment of the magnet? Solution :  Magnetic field strength , B = 0.25T , Torque on the bar magnet , $T = 4.5 \times 10^{-2}J$ ,  Angle between the bar magnet and the external magnetic field , $\theta = 30^{\circ}$. then Torque is related to magnetic moment (M) as : $$\tau = MBsin\theta$$ $$\therefore M = \frac{\tau}{Bsin\theta}$$ $$M = \frac{4.5 \times 10^{-2}}{0.25 \times sin30^{0}}$$ $$M = 0.36 JT^{-1}$$ hence , the magnetic moment of the magnet is $M = 0.36 JT^{-1}$ 5.2 A short bar magnet of magnetic moment $M = 0.32 JT^{-1}$ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable. and (b) un...

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 , $$\overrightarrow{B_{2}} = \frac{\mu_{0}}{4 \pi}\frac{q_{m}}{(r+l)^{2}}(-\hat{i})$$ $\therefore$ Net magnetic field intensity at a point P due to the bar magnet. $$\overrightarrow{B...

5 Laws of Friction  - Class 11 Physics - Laws of motion - Param Himalaya Laws of friction are stated as :   1. The direction of the limiting force of friction is always opposite to the direction in which the motion of a body takes place or tends to take place. 2. The force of friction acts tangentially along the surfaces of contact of two bodies. 3. The magnitude of the limiting force of friction (f) is directly proportional to the normal reaction (R) between the two surfaces , that is , $f \propto R$  4. Limiting force of friction is independent of the area of the surfaces of contact of two bodies as long as normal reaction is constant. 5. Limiting force of friction depends on the nature of the surfaces in contact.

Angle of Friction : The angle between the normal reaction and the resultant of limiting force of friction and normal reaction is called angle of friction. The resultant of limiting friction (F) and normal reaction (R') is represented by R = OC. the angle $\theta$ between R and R' is called angle of Friction. Now $tan \theta = \frac{(f_{s})_{max}}{R} = \frac{F}{R}$ But $\frac{F}{R} = \mu_{s}$(co-efficient of static friction) $\therefore tan\theta = \mu_{s}$ Thus , co-efficient of static friction is numerically equal to the tangent of the angle of friction.

NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism 4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil? Solution :   Given , n = 100 , r = 8 cm = 0.08m , I = 0.4 A  Magnitude of the magnetic field at the centre of the coil is given by the relation, $$\left | B \right | = \frac{\mu_{0} }{4\pi }.\frac{2\pi nI}{r}$$ where $\mu_{0} =4\pi \times10^{-7}TmA^{-1}$ $$B = \frac{4\pi \times 10^{-7}}{4\pi }.\frac{2\pi \times 100 \times 0.4}{0.08}$$ $$B = 10^{-7}. \frac{2\pi \times 10^{2} \times 4 \times 10^{-1} }{8 \times 10^{-2}}$$ $$B = \frac{2\pi \times 4}{8}.\frac{10^{-7} \times 10^{^{2}} \times 10^{-1}}{10^{-2}}$$ $$B = \pi \times 10^{-7+2-1+2}$$ $$B = 3.14 \times 10^{-4}T$$ 4.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire? Solution :  Given I = 35 A , r = 20 cm ...

Ampere's Circuital Law :   Ampere Circuital Law states that the line integral of the magnetic field around any closed path in free space is equal to absolute permeability ($\mu_{0}$) times the net current passing through any surface enclosed by the closed path.  Mathematically :  $$\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{0}I$$ Where , $\overrightarrow{B}$ is the magnetic field , $\overrightarrow{dl}$ is the small element , $\mu_{0}$ is the absolute permeability of free space and I is the current enclosed by the closed path. Proof : Consider an infinitely long straight conductor carrying current I. The magnetic field lines are produced around the conductor as concentric circles. The magnetic field due to this current carrying infinite conductor at a distance a is given by $$B = \frac{\mu_{0}}{4\pi }(\frac{2I}{a})$$                            Consider a circle of radius a around the wire (cal...

Biot-Savart's law is used to determine the strength of the magnetic field at any point due to a current carrying conductor. Consider a very small element AB of length dl of a conductor carrying current I. The strength of magnetic field dB due to this small current element $(I\overrightarrow{dl})$ at point P, distant r from the element is found to depend upon the quantities as under: $$(i) dB \propto dl$$ $$(ii) dB \propto I$$ $$(iii) dB \propto sin\theta$$  where $\theta$ is the angle between $\overrightarrow{dl}$ and $\overrightarrow{r}$. $$(iv) dB \propto \frac{1}{r^{2}}$$ Combining (i) to (iv) , we get  $$dB \propto \frac{Idlsin\theta}{r^{2}}$$ $$dB = k \frac{Idl sin\theta}{r^{2}}$$ Where k is a constant of proportionality. In S.I units, $$k = \frac{\mu_{0}}{4\pi}$$ where , $\mu_{0}$ is called absoulte permeability of free space i.e . vacuum. Hence , equ (i) becomes  $$dB = \frac{\mu_{0}}{4\pi}. \frac{Idlsin\theta}{r^{2}}$$ Value of $\mu_{0}$ in S.I units = $4\pi \time...