Class 12 Physics CBSE Magnetism and Matter PDF

Chapter 5 Magnetism and matter Magnetic field lines These are the imaginary lines which are used to represent the magnetic field... # Properties of magnetic field lines The direction of magnetic field lines outside the magnet is from North P...

Chapter 5 Magnetism and matter Magnetic field lines These are the imaginary lines which are used to represent the magnetic field... # Properties of magnetic field lines The direction of magnetic field lines outside the magnet is from North Pole to South Pole. And from South Pole to North Pole through the magnet. It forms a closed loop is Tangent of the magnetic field line at any point gives the direction of magnetic field at that point.  The magnetic field lines never intersect. If two lines intersect at any point, two tangents can be drawn at that point, which showing two direction of field. This is not possible. The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field B. & In a uniform magnetic field, magnetic field lines are parallel and they are equally spaced.  CBSE 2023 Question Explain why do magnetic lines of force form continuous closed loops. > - Dipoles & i S & B ds. = O - & Magnetic dipole moment m = NIA It is the product of pole strength and distance between the poles. The magnetic dipole moment m is a vector quantity whose direction is from South Pole to North Pole of the magnet. c 2a- [m2a - m = NX S Polestrength 29m - 2- > length of magnet Question ↓ A bar magnet is cut into two equal pieces transverse to its length. - What happens to its dipole moment? 9m = am Am 9m M = 29qm 2a- A mi = 99m = m = M/z  an Am 9m N am-Am/ = m 29 m = M/s = The electrostatic analog grets Eurostatics Beat e · E · NS 0- Bac = Bar magnet as an equivalent solenoid Solenoid and bar magnet produces similar magnetic effects. Therefore we can say that bar magnet is an equivalent solenoid. · Bax N S -- Baxi ↳I B Bea : Lo n s  ~ The dipole in a uniform magnetic field If a magnetic needle is placed in a magnetic field and left, it will experience a torque and start oscillating in simple harmonic motion. The torque on the needle is T = mBSino = mXB Time period of oscillations of magnetic needle Ta Is moment of inertia of banmagent F Potential energy of a magnet in a magnetic field U = -m. Bw - mBlosO Question 0 = 0 A bar magnet has magnetic dipole moment M. Its initial position is parallel to the direction of uniform magnetic field B. In this position, the - magnitudes of torque and force acting on it, respectively are CBSE 2022 (Term-1) (a) 0 and M B MB Sino (b) M B and M B T = (c) 0and 0 EMXBX O (d) |MxB| and 0 Force X = o Torque - Eo -  Non uniform m F Rotation of translati. - - Fore # N S torque  w S Stable abi S > - N Angle bur Question S - Bam = d The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform magnetic field is (dipole moment of this dipole = 0.5 A m2) Uz-mB losR (a) 0.07 J (b) 0.08 J (c) - 0.09 J (d) 0.1 J Unstable - Work done change in p E. N S = N Y S = PEX-P. Emi S --mBCosIso--mB(o0 Bam Of 180 - - mBX -1 - - mBX) - mB + mB - = GmB = 2x0 09X. 0. 5 = 5  Sin Question What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5 cm at a distance of 50cm from the midpoint? The magnetic moment of the bar magnet is 0.40 Am². MOT T Bea- Band : 2 Ben Question CBSE Delhi Outside 2020 2 A bar magnet of dipole moment 3 Am rests with its centre on a frictionless pivot. A force F is applied at right angles to the axis of the magnet, 10 cm from the pivot. It is observed that an external magnetic field of 0.25 T is required to hold the magnet in equilibrium at an angle of 30° with the field. Calculate the value of F. How will the equilibrium be effected, if F is withdrawn? Turque due to MF Turque due to extend fore  t= rxF Net terque = zero MF Torque due to external feed Torque due to = MB Sno = V Sinoe , 3 x0 25 y Sinso. = 0. 1x F Sin 90 25 = 0.xFx = F Gauss law in magnetism The surface integral of magnetic flux over a closed surface is zero. If B is the magnetic field intensity on a closed surface then, Go B. ds-o Magnetic monopole. does not exist  6t. ds = /Er Question If magnetic monopoles existed, how would the Gauss’s law of magnetism be modified? G B. ds-Mom  Bon Mo E B = Terms related to magnetism mater 1.Magnetisation in a of bringing up magnetism > - process 2.Intensity of magnetisation (M) = 3. Magnetic filed intensity (B) or Magnetic flux density B= H 4. Magnetic intensity (H) or Magnetising field H = nI 5. Magnetic susceptibility (χ) How it respond Xm = 6. Magnetic permeability (μ) = M= Relation between B, M and H B = μ₀ [H + M] Relation between relative permeability and susceptibility μ = 1+ χ z m Question Obtain the relation connecting relative permeability and magnetic susceptibility Total MF ME due to solenced + mF due to metend essa L = B = MoUI + MoM B = MoH + MoM = B= Mo[H +m] - with If & = Mo + MoXm M = Mo [1 + Xm] ( Xm) # + = Mr = It Xm  Magnetic properties of materials Magnetic substances are divided with respect to their magnetic properties into weakly magnetic and strongly magnetic substances.  According to magnet properties magnetic materials are classified into three. Magnetic material - - Diamagnetic Paramagnetic Ferromagnetic Au , Ag , Cu, Al, Na, Mg, Li, Ti, Pt, Iron, cobalt, H₂O , Hg , Zn Liquid oxygen nickel  Diamagnetic Paramagnetic Ferromagnetic Weakly Weakly Strongly Magnetisation magnetised in the magnetised in the magnetised in the opposite direction direction of direction of of applied field applied field applied field [] S 1] (m50-  Diamagnetic Paramagnetic Ferromagnetic Relative permeability and χ is small and χ is small and M χ and μ are very M z M negative positive high susceptibility -1