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FLECTRO MAGNETIC INDUCTION MAGNETIC FLUX In analogy with the electric flux $., a magnetic flux $,, of the magnetic field for a surface is defined. Imagine dividing a mathematical surface into infinitesimal area elements. The direction of an area element da ata point on the surface is perpendicular to the surface at that point. A typical element for a surface is shown in fig. along with the magnetic field B at a point. eq The magnetic flux for a surface is The magnetic flux for an infinitesimal proportional to the number of field area da is given by dd = B, da lines intersecting the surface. > The magnetic flux do for the area element "da" is dp=B . da 3 If the magnetic field B makes an angle 4 with the normal to the surface as shown in fig. then the normal component of the field is B cos 4 and in this case, the magnetic flux is given by 6=BAcosd or $=B. A A ay = BA cost noe = BA Onin = 0 So, magnetic flux linked with a closed surface may be defined as the product of the surface area and the normal component of the magnetic field acting on that area. It may also defined as the dot or scalar product of magnetic field and surface area. The magnetic flux for a general surface is obtained by integrating (summing) the contributions do as the area element da ranges over the surface. Thus, o=/ B.da Physically it represents total lines of induction passing through a given area. EMI POSITIVE AND NEGATIVE FLUX In case of a body present in a field, either uniform or non-uniform Out ward flux is taken to be positive while inward negative. B Total flux = 0 Nonuniform-field Total flux = 0 Uniform-field If the normal drawn on the surface is in the direction of the field, then the flux is taken as positive. In this case, 6 is 0°. or 8 < 90° then the flux is taken as positive. If the normal on the surface is opposite to the direction of the field, then @ = 180°. In this case, the magnetic flux is taken as negative. Magnetic flux density, B -2 Lines of force are imaginary, but as magnetic flux associated with elemental area da ina field B, so flux is a real scalar physical quantity ()-@lbl-[F lel tas F = BIL sino AL 2 DIMENSIONS : [ML2T2A-1] lity ws | (2]=IMUT A UNIT : SI volt * sec. As [ML?T~] corresponds to eneray, It ls known as weber (Wb) SI unit of magnetic flux will be or T.m? (as tesla = Wb/m?) joule joulexsec [as anipere= soso ampere coulomb CGS maxwell (Mx) = volt sec [as (joule/coulomb) = volt] 1Wb=1V xs = 10° emu of pot". x s = 10° Mx [1 volt=10° emu of potential] Special Note : $5 - |B .da Soif B=0, then 6,=0. > But if 6,=0, B may or may not be zero. Because if B = 0, flux for some elements may be positive while for the other negative giving zero net flux, e.g., for a closed surface enclosing a dipole $, = 0 but db and hence B<0. Ex. Ata given place, horizontal and vertical components of earth's magnetic field B,, and B, are along x and y axes respectively as shown in fig. What is the total flux of earth's magnetic field associated with an area A, if the area A is in (a) x-y plane (b) y-z plane and (c) z-x plane? Sol. ~ B =iBy—jBy = const. and o=(8 _da-B-A [as B= const.] i 8. 8. Bf sal a, Ar A x + x +x B, Lo B, 2 zZ Zz for area in x-y plane : A=Ak by = (IBY — iBy). (KA) =0 [v i.k=j.k=0] for area Ain y-z plane : A=Ai by: = (iBy —jBy). (7A) =ByA [« i.7=1 and j.j=0] for area S in z-x plane: 4— Aj 62, = (iB ~jBy). (JA)=-ByA [ i.j=0 and j.j=1] Negative sign implies that flux is directed vertically down. DIFFERENT WAYS WHICH CAN VARY THE MAGNETIC FLUX : Magnetic flux in planar area A due to an uniform magnetic field B ¢ = BA cosd of these changes. (1) By varying the magnetic field B with time Due to motion of magnet B will change with time. Or if magnetic field produced due to current carrying loop then due to change in current change in B will occur with time. (2) By varying the area of the conducting loop A with time: > FOO Special Note: Ifanyoneofthese v ||B, / ||B, ¢ || v then induced e.m.f. is zero. Ex. An artificial satellite with a metal surface has an orbit over the equator. Will the earth's magnetism induce a current in it ? Explain. Sol. The satellite will cut vertical component in the equatorial plane is zero. Consequently there will be no change in magnetic flux and hence no current will be induced. Ex. A conductor of length 10 cm is moved parallel to itself with a speed of 10 m/s at right angles to a uniform magnetic induction 10 Wb/m?. What is the induced e.m.f. in it? Sol Given: ¢ = 10 cm = 0.1 m, v = 10 m/s B = 10* Wb/m? e.m.f. induced in conductor e=BfV=10*4*0.1* 10=10¢7V Ex. A horizontal telegraph wire 10 m long oriented along the magnetic east-west direction falls freely under gravity to the ground from a height of 10 m. Find the emf induced in the wire at the instant the wire strikes ground. (B, = 2.5 * 10° Wb/m?, g = 9.8 m/s?) Sol. Given: u = 0, a = -g = -9.8 m/s’, s = -10 m, B, = 2.5 * 10° Whim’, €=10m Let the speed of wire at last moment = v then w?=0+ 2 (-9.8) = (-10)=196 [wWe=u?+2as] A v= 14 m/s e.m.f. induced in wire e = BL, fv = 2.5 x 10% * 10 * 14=3.5 * 10° Volt Ex. Awire kept along north-south is allowed to fall freely. Will an induced emf be set up? Sol. No. This is because there will be no change of magnetic flux in this case. 12 Ex. A metal block and a brick of the same size are allowed to fall freely from the same height above the ground. Which of the two would reach the ground earlier and why? Sol. The brick will reach the ground earlier. This is because the eddy currents produced in the metal block will oppose the motion of the metal block. Ex. A jet plane is travelling west at the speed of 1800 km/h. What is the voltage difference developed between the ends of the wing 25 m long if the earth's magnetic field at the location has a magnitude of 5.0 x 10“ Tesla and the dip angle is 30°? 1800 «1800 Sol. Given that v = 1800 km/h = ~B0x60 m/s = 500 mis €=25 m, B = 5.0 * 10“ T and 6 = 30° Vertical component of earth's magnetic field B, = B sin 0 = 5.0 x 10+ = sin 30° = 2.5 x 104 T MOTIONAL e.m.f. Now suppose the moving rod slides along a stationary U-shaped conductor, forming a complete circuit. No magnetic force acts on the charges in the stationary U-shaped conductor, but there is an electric field caused by the charge accumulations at a and b. Under the action of this field a counterclockwise current is established around this complete circuit. The moving rod becomes a source of elec- tromotive force. Within it, charge moves from lower to higher potential and in the remainder of the circuit, charge moves from higher to lower potential. We call this a motional electromagnetic force denoted by e, we can write, e® 8® @ 8 eye @/] ® @| @)@)] @ e® @ 8° ®@ e=Bvi If R is the resistance of the circuit, then current in the circuit | i= eo ® ® ® ® a e[e® @ ® ® 1 ‘ e= Bw : é= ®@ @ ® ® = ns é 2 @ b eS g In the figure shown, we can replace the moving rod ab by a battery of emf Bvl with the positive terminal at a and the negative terminal at b. The resistance r of the rod ab may be treated as the internal resistance of the battery. Hence, the current in the circuit is, D <= = e Rer or i a + = 13 Induction and energy transfers : In the figure shown, if you @ @ @ @ ® move the conductor ab with a constant velocity v, the current in a z ; @ @ a 8 8 the circuit is, Jr 2 Fv Bye @ o. 9 @ @ i= RR (r= 0) = @ @ b @ @ 252 Amagnetic force F_=i/ B= acts on the conductor in opposite direction of velocity. So, to R move the conductor with a constant velocity v an equal and opposite force F has to be applied in the conductor. Berty Thus, F=F_= us Le R Bir?y? The rate at which work is done by the applied force =P... =Fw= R a low DD a a 
