Tiêu đề ELECTROMAGNETICS INDUCTION.pdf ✅

 Digital www.allendigital.in [ 119 ] Magnetic Flux The magnetic flux () linked with a surface held in a magnetic field (B) is defined as the number of magnetic field lines crossing that area (A). If  is the angle between the direction of the field and normal to the area, (area vector). A plane of surface area A placed in a uniform magnetic field B . No. of magnetic field lines passing from any surface area is called magnetic flux.  =  =  B A BAcos If coil has N turns,  =  NBAcos where,  → Angle between A and B Flux Linkage If a coil has more than one turn, then the flux through the whole coil is the sum of the flux through the individual turn. If the magnetic field is uniform, the flux through one turn is  = BAcos So, for N turns, the total flux linkage  = NBA cos Note: (1) Magnetic field lines are imaginary, magnetic flux is a real scalar physical quantity with dimensions (2) Magnetic flux is a scalar quantity (3) Units: (a) SI Unit : Weber (Wb) (b) Derived SI unit : Tesla-meter2 (T-m2) (c) CGS UNIT : Maxwell (Mx) (d) Conversion factor : 1 Wb = 108 Mx Dimensional formula of magnetic flux  = [M L2 T–2 A–1] (4) If magnetic field is non-uniform than magnetic flux is given by  =  B dA   BሬԦ AሬԦ 02 Electromagnetic Induction
NEET : Physics [ 120 ] www.allendigital.in  Digital (5) Gauss law in magnetism: Net magnetic flux through closed surface is always zero.  =  = B dA 0  • Since incoming field lines = Outgoing field lines • Net flux is zero. • Magnetic monopoles does not exist. (6) Rate of change of flux is given by (a) Instantaneous Rate d dt  = (b) Average Rate Final Ininal t t   −  = =   Illustration 1: A rectangular loop of area 0.06 m2 is placed in a magnetic field 1.2 T with its plane inclined 30° to the field direction. Find the flux linked with plane of loop. Solution: Area of loop A = 0.06 m2, B = 1.2 T and  = 90° – 30° = 60° So, the flux linked with the loop is  = BAcos = 1.2 × 0.06 × cos60° = 1.2 × 0.06 × 1/2 = 0.036 Wb Illustration 2: A loop of wire is placed in a magnetic field B 0.3jT = ˆ . Find the flux through the loop if area vector is ˆ ˆ ˆ 2 A (2i 5 j 3k) m = + − Solution: ˆ ˆ ˆ B (0i 0.3 j 0k)T = + + Flux linked with the surface ˆ ˆ ˆ ˆ  =  =  + − B A (0.3 j) (2i 5 j 3k) T-m2 = 1.5 Wb T-m2 = Wb Illustration 3: At a given plane, horizontal and vertical components of earth's magnetic field BH and BV are along x and y axes respectively as shown in figure. What is the total flux of earth's magnetic field associated with an area S, if the area S is in (a) x-y plane (b) y-z plane and (c) z-x plane. Solution: H V B iB jB = − ˆ ˆ = constant, so  =  B S [ B = constant] (a) For area in x-y plane ˆ S Sk = xy H V ˆ ˆ ˆ  = − = (iB jB ).(kS) 0 (b) For area S in y-z plane ˆ S Si = yz H V H ˆ ˆ ˆ  = − = (i B jB ).(iS) B S (c) For area S in z-x plane ˆ S S j = zx H V V ˆ ˆ ˆ  = − = − (iB jB ).(jS) B S Negative sign implies that flux is directed vertically downwards. BV BH S x z y (a) Bv BH S x z y (b) BV S BH x z y (c) N S
Electromagnetic Induction  Digital www.allendigital.in [ 121 ] Illustration 4: If a coil of area A 3i 4j = + m is placed in magnetic field B 2i 3j 4k = + + T. Then find the flux passing from this coil. Solution:  =  B B A = (2i 3j 4k) (3i 4j) + +  + = 6 + 12 = 18 T-m2 or 18 Wb Illustration 5: Circular coil of 500 turns & area 5 cm2 is placed in uniform magnetic field B = 2T such that its area vector makes an angle of 60° with the direction of field then find flux passing from this coil? Solution: Given: A = 5 cm2 = 5 × 10–4 m2, N = 500, B = 2T,  =  60  =  NBAcos  ( )( )( ) 4 500 2 5 10 cos60 −  =     = 0.25 Wb Illustration 6: The change in flux through the ring of area ‘A’ if it is rotated by 180° in uniform magnetic field(B) as shown: Solution:  =  − 2 1 =BAcos BAcos 2 1  −   −  BA cos0 cos180 ( )  2BA Illustration 7: Find magnetic flux passing through this square loop. Solution: At any distance x from wire, flux of small area is given as d B dx = ( ) 0 I dx 2 x  =   t 0 t I dx 2 x +   =     t 0 t I dx 2 x +   =    0 I t n 2 t    +  =      BሬԦ AሬԦ I t  x dx
NEET : Physics [ 122 ] www.allendigital.in  Digital Illustration 8: Given coil is placed in external magnetic field then find flux through coil. Solution:  =  NBAcos {Here  is angle between B & A }  =  NBAcos90  = 0 Illustration 9: Given coil is placed in external magnetic field then find flux through coil. Solution: Here angle  between B & A is 90° So, flux  =  = NBAcos90 0 1. A coil of 100 turns, 5cm2 area is placed in external magnetic field of 0.2 Tesla (S.I.) in such a way that it makes an angle 30° with the field direction. Calculate magnetic flux through the coil (in weber). 2. A coil of N turns, A area is placed in uniform transverse magnetic field B. If it is turn through 180° about its one of the diameter in 2 seconds. Find rate of change of magnetic flux through the coil. 3. A square cube of side 'a' is placed in uniform magnetic field 'B'. Find magnetic flux through each face of the cube. 4. The magnetic field perpendicular to the plane of a loop of area 0.1 m2 is 0.2 T. Calculate the magnetic flux through the loop. 5. The magnetic field in a certain region is given by ( ) ˆ ˆ B 4i k = − tesla. How much magnetic flux passes through the loop of area 0.1m2 in this region if the loop lies flat in xy plane? 6. A solenoid 4cm in diameter and 20cm in length has 250 turns and carries a current of 15A. Calculate the flux through the surface of a disc of 10cm radius that is positioned perpendicular to and centered on the axis of the solenoid. AሬԦ BሬԦ I BEGINNER’S BOX-1 G H D C B E A F B a