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.