Title Moving Charges And Magnetism New Notes.pdf ✅

Sunil Jangra P hysics G ru ij u CH-4th Moving Charges and Magnetism Class 12 Physics Magnetic Field:-the magnetic field is the space around a current carrying conductor or space around a magnet in which its magnetic effectcan be felt. Denoted by 5 and SI unit is Tesla (T). · Magnetic force on a moving charge The magnetic force (Em) on a charge a moving with velocity in a magnetic field B is given, both in magnitude and direction -> by Fm = 9(4x5) In magnitude Fm=Bqusino - Note: Em=0 Fiit8 where O is the angle between and B. Force is maximum when 0-90' if 4= 0 minimum when O= 0 or 180 inB or MMB Direction ofImo- I is perpendicular to both and 5, and given by Right handrule. if 9 is positive then Fm'44YX5 if 9 is negative then -> Fr 4NTXB Rules toFind direction ofMagnetic Field and Direction of Force · Right Hand Thumb Rule Right Hand Rule Ea a =YXB Right-hand rule Y i x B, * Yat L · Right Hand Palm Rule. Fleming leftHand Rule · By convention the direction ofMagnetic Field isperpendicular to the paper going inwards is shown by and direction perpendicular to the paper coming out is shown by 0. [prectionof perpendicular [DirectionatBeateninlan Definition ofB if v = 1,q= 1 and sinc-1 or 0=90 then from De F = IXXB = B i.e the magnetic field induction or magnetic flux density at a point in the magnetic field is equal to the force experienced by a unit charge moving with a unit velocity perpendicular to the direction ofmagnetic field at that point. unit of: SI unit of Magnetic field is Tesla (T) or weber/metre i.e Wblm? or NSCm! we know thatF= quBsino and B = sino crys = ITesla. Dimension ofB = I=> SMAY
Sunil Jangra P hysics G ru ij u · path of a charged Particle in Uniform Magnetic field. - · When 1 B then Here Centripetal force - Magnetic force 2 F = 9VB, This magnetic Force MX = 9VB is perpendicular to the velocity at every instant. = lence path is circle. v = MX P=M SO v= 1 B aB and P-momentum = 1km K= kinetic energy so v = m and K=9X aB Time Period of circular path T= ctr or T= 2 Fimi angular speed a= GI = Ba angra and frequency ofrotation f = f= 14- Bu · When I makes an angled with 5. In this case velocity can be resolved in two components (i.e along 5 and perpendicular to 5.) Y1 = x0s0 & X= Vsin NOTE: YI gives a circular path and y gives straight line path. The resultant path is a helix as shown in figure. The radius of this helical path is v = mY1 = mssino --- B9 and ↑ine Period T= 2IM and =Ba aB 21m Pitch- Pitch is defined as the distance travelled along magnetic field in one complete cycle. ive P= YIT = P= CVC00)GIIm · Magnetic Force on a Current carrying conductor Suppose a conducting wire carrying a current i, is placed in a magnetic field. 5. The length of wire is I and area of cross section is A. The Free electrons drift with a speed Va opposite to the direction ofcurrent. The magnetic force exerted on the electronisdEmE-eCTXT= - eCGAITax) where n = number of free electron per unit volume and m = i(Ex) where HeAVd=i of the wire. Biot Savart Law- This law gives the magnetic field due to an infinitesimally small current carrying wire at a some pointerfa the magnetic field 65 at a point passociatedwithan length element a ofa wire carrying a steady current i. S d5= Moi (dix) where Y=r and 02 I Mo=187tm KIT Magnitude of 65 is Id5 = no sinin SNOTE CT: which points in the direction ofCurrent 3.
Sunil Jangra P hysics G ru ij u " Direction of dB→ : dB→↑↑(dixit , so dB→ is along di✗ F and dB→=0 , if di is in the plane of Paper. Here r = tin • Magnetic field on the axis of a circular current loop. Acc to Biot Savart's Law- dB = Me idlsin900 Because 13 is very small , 4't F therefore angle between di & F is 90° . so dB = us idle - 1. 41T I R4nY This dB is perpendicular to plane ° ' formed by di' and F I. e (di ✗ F) dBn= DBCOSO dB is resolved in two component dBA and DBL. Here summation of dB, over the complete cycle is zero. so magnetic field is due to only dB. Component. i. e B = / DBCOSO = / Us idle COSO Here 1050=11 41T (R4Ñ) (124×742 For N Turn 13=4%-1%4!gTμ+ñ,k㱺B=¥Y¥㱺%É" 㱺 B = noir " "TR 㱺 B%;÷%j% B-_ 41T ( 122+25312 ¥;¥%% • agnatic Field at the centre of a current loop. so 2=0 B = No Niti 㱺 B. = Noni 2123 -212 Ampere's circuital Law : The line integral of magnetic field induction B- around any closed path in vacuum is equal to no times the total current threading the closed path i.e § B→ . di = Alo I At every point of the closed path B-' II di. • Ampere's circuital Law holds good for a closed path of any size and shape around a current carrying conductor because the relation is independent of distance from conductor. Application of Ampere's circuital law solenoid : A solenoid is a closely wound helix of insulated copper wire. U • All the magnetic field lies within the core of solenoid i.el uniform. • Outside the solenoid there is no magnetic Field. Magnetic Field due to solenoid Figure A shows a longitudinal cross-section of part of such a solenoid carrying current i. since solenoid is ideal B- in the interior is uniform and parallel to axis and 15 in the exterior is 0. consider the rectangular path of length 1 and width H. Now apply Ampere's circuital law to this path Fig (A) III. diya 0 : 13=0 ftp.dipxo.aa-o % B→-1 di kit and /&- di) = BI = NONE N=Total no. of Turns ab n= no. of turns per ← I → B-- Mo Nyi unit length. B-- none n=^{
Sunil Jangra P hysics G ru ij u Toroid: A toroidal solenoid is an anchor ring around which is large number of turns of a copper wire are wrapped. A toroid is an endless solenoid in the form of a ring. • Magnetic Field due to Toroid dashed lines are Figure shows a cross-sectional Amperian loop . View of toroidal solenoid. • Amperian loop 1 encloses no current i.e inet=o so 13=0 • For Amperian loop 3 , the current coming out of the plane of the paper is exactly by the current going into it. • Now consider the Amperian loop 2 , then § B?di= No inet 㱺 13121in) = NoNi or B = Modi 211-8 N_ = n (number of turns per unit length of torus). B = Noni 21Tr Force Between Parallel Current Carrying wires consider two long wife a and b kept parallel to each other at a distance d and carrying currents ia and ib resp. in the same direction Magnetic field on wire b due to current Ia is BE No ia ( In 4 direction) 2Nd Magnetic force on a small element L of wire b due to this magnetic field is EE ib LBa 㱺 F- = ibllloia 2Nd Fbi Force on b due to a Force per unit length of wire b due to wire a % Fga = No iaib Similarly Fab = No iaib 2'tr T 2 ITV NOTE:→ The wires attracts each other it currents in the wires are flowing in same direction and they repel each other if the currents are in opposite directions. Magnetic Dipole :→ Every current carrying loop in a magnetic dipole . it has two poles South t.SI and North CNI • This is similar to a bar Magnet. 9T is denoted by M. = iA→ > Magnetic dipole moment is a vector ✗ ATni a r quantity. it's direction from > < South to North Pole. • Torque on a current loop placed in Uniform Magnetic Field. Case-1 when Ñ 1- B→ There are no force on arm AD and CB ° : length of arm Parallel to B→• And force on arm AB and DC are equal and opposite. Hence to net force on the loop is zero. So Torque is there. Torque=/ either Force / ✗ 1- distance between Forces. i.e I = IbB ✗ a 㱺 I = IAB A- = Area of rectangular loop i.e A= ba gwviljaipr • case-2 : when Ñ makes an angle 0 with B The forces on arm AB and KD are equal and opposite • i.e Net Force is zero.