Chapter 29--ExamplesProblemFrom Example 29.3SolutionPart BSlide 6Slide 7EMF=-DF/Dt=-B*DA/DtSlide 9Case a—Rotating about y-axisCase b—rotating about the x-axisCase c—Rotating about z-axisProblemMagnetic Field of a Circular LoopEMF in smaller loopDirection of current1Chapter 29--Examples2Problema) Derive the equation relating the total charge Q that flows through a search coil (Conceptual Example 29.3 in Section 29.2) to the magnetic field B. The search coil has N turns, each with area A and the flux through the coil is decreased from its initial value to zero in a time t. The resistance of the coil is R and the total charge is Q=I t where I is the average current induced by the change in flux3From Example 29.3A search coil is a practical way to measure magnetic field strength. It uses a small, closely wound coil with N turns. The coil, of area A, is initially held so that its area vector A is aligned with a magnetic field with magnitude B. The coil is quickly pulled out of the field or rotated.Initially the flux through the field is =NBA. When it leaves the field or rotated,  goes to zero. As  decreases, there is a momentary induced current which is measured. The amount of current is proportional to the field strength.4SolutionEMF=-/t–Where =NBA and =0EMF=NBA/tV=iR where V=EMFEMF=NBA/t=iRi=NBA/RtQ=it=NBA/R5Part BIn a credit card reader, the magnetic strip on the back of the card is “swiped” past a coil within the reader. Explain using the ideas of the search coil how the reader can decode information stored in the pattern of magnetization in the strip.6SolutionThe card reader contains a search coil.The search coil produces high EMF and low EMF as the card is swiped.A high EMF is treated as a binary 1 and a low EMF is treated as a binary 0Ascii information (character codes between 1 and 64) can be stored there in a few bits (6-bits or 26).7ProblemA circular loop of flexible iron wire has an initial circumference of 165 cm but its circumference is decreasing at a rate of 12 cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop and with a magnitude of 0.5 Ta) Find the EMF induced in the loop at the instant when 9 s have passed.b) Find the direction of the current in the loop as viewed looking in the direction of the magnetic field.8EMF=-/t=-B* A/ t=BAd/dt=BdA/dtc=2r so A=r2r=c/2 so A=c2/4dA/dt=d/dt(c2/4c/4)dc/dtd/dt= Bc/2)dc/dtWhere B=0.5dc/dt=0.12 m/sAt 9s, c=1.65-.12*(9s)=0.57 md/dt=5.44 x 10-3 VSince  is decreasing, the EMF is positive. Since EMF positive, point thumb along A and look at fingers.They curl counterclockwise.9ProblemSuppose the loop in the figure below is a) Rotated about the y-axisb) Rotated about the x-axisc) Rotated about an edge parallel to the z-axis.What is the maximum induced EMF in each case if A=600 cm2, w=35 rad/s, and B=.45T?10Case a—Rotating about y-axisVBAEMFtBAtBAdtdtbutBAdtdABdtddtdEMF945.035*06.0*45.0sincoscosmax11Case b—rotating about the x-axisThe normal to the surface is in the same direction as B during this rotation. Thus B-A=constantd/dt(constant)=0 so EMF=012Case c—Rotating about z-axisVBAEMFtBAtBAdtdtbutBAdtdABdtddtdEMF945.035*06.0*45.0sincoscosmax13ProblemThe figure below shows two parallel loops of wire having a common axis. The smaller loop (radius r) is above the larger loop (radius R) by a distance x>>R. Consequently, the magnetic field due to current i in the larger loop is nearly constant throughout the smaller loop. Suppose that x is increasing at a constant rate, v. 1) Determine the magnetic flux though area of the smaller loop as a function of x.2) In the smaller loop find1) The induced EMF2) The direction of the induced current.xRadius=rRadius=Ri14Magnetic Field of a Circular Loop 32023222022xiRRxiRBBy the RH Rule, the field is upward in the small ring2320232022rxiRABrAandxiRB15EMF in smaller loop42204332202320232032311222xvriREMFvdtdxbutdtdxxxdtdxdtdriRrxiRdtddtdEMFrxiRAB16Direction of currentAssume that normal to smaller loop is positive upwardsB, then, is in positive directionBut  is decreasing or has a negative d d /dtA negative * negative=positive so EMF is +RH Thumb in up direction, fingers curl counter