exam 2 physics

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Physics

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Apr 3, 2024

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-A charged particle is launched with a velocity of 5.2 x 10° m/s at an angle of 35°...determine the magnitude of the charge on the particle. F=q*v*B*sin(8), g=F/ v*B*sin(0) =0.0026N/(5.2*10°m/s)*(0.0045T)Sin(35°)=19x10°C= 19uC -An electron is moving with a speed of 3.5 x 10° m/s...What is the magnitude of the magnetic force on the electron? F=q*v*B*sin(0)= (1.6*10°C)(3.5*10°m/s)(0.60T)(sin60°)=2.9098 x 10"* N -A particle with a mass of 6.64 x 10?” kg and a charge of +3.20 x 107" C is accelerated what is the magnitude of the magnetic force exerted on the particle? %*m*v?=q*Av => v?=3.2x109C*2.45*10°V / %*6.64*10 ?’Kg=1.5367*10"m/s, v=1.5367*10’m/s, Mag Forc=F=q*v*B*sin(90)= 3.20*10*°C*1.5367*10’m/s*1.60T*1= F=7.87 x 10-12 N -A mass spectrometer is used to separate two isotopes of uranium with masses m; and m; ... B e/ what is the value of the ratio mi;/m2? K=q.v & k=mv?/2 => m=qB?*r?/2v=>(v,q & B are the same for both) m1/m2=(r1)?/(r2)?*= Ratio=0.9874 -A 0.0150-m wire oriented horizontally between the poles of an electromagnet carries a direct current of 9.5 A ..... what is the magnitude and direction of the magnetic force on the wire between the poles? : F=I*L*B*sin6=>9.5A*0.015m*0.845T*sin(25°)=0.0509N, upward -A long, straight wire carries a 6.0-A current that is directed in the positive x direction. When a uniform magnetic field is applied perpendicular to a 3.0-m segment of the wire, the magnetic force on the segment is 0.36 N, directed in the negative y 7 direction, as shown. 2! L. What are the magnitude and direction of the magnetic field? current=1=6iA Length=L=3m Mag Force=F=-0. 36]N -A loop of wire with a welght of 0.55N is orlented vertically and carrles acurrent | =2.25 A. ... The net force on the wire is il measured using a balance and found to be zero. - What is the magnitude of the magnetic field? Fb=1*L*B => B=Fb/I*L equate Fb with weight of wire 0.55N=>B=0.55N/2.25Amp*0.2m=1.2T -A circular coil consists of 5 loops each of diameter 1.0 m. The coil is placed in an external magnetic field of 0.5 T. Determine the angle between the normal to the plane of the coil and the direction of the magnetic field. T=B*Area*N*I*sin6, Area=nt*(r)>=n*(0.5)’=n/4, sin0=3.93Nm/(0.5T)(r/4)(5)(4Amp) =0.50038< sin™ (0.50038)= Angle is of 30° -A single circular loop of radius 1.00 m carries a current of 10.0 mA. It is placed in a uniform magnetic. [ 0o mA PN N, magnitude of the torque exerted on the loop by the magnetic field? TnMN‘i‘B‘A‘(san), Area=n*r’=n*12=m, Torque=(1)(10x107%A)(0.5T)()(1m)? (sin90° is 1)=15.7x10°N.m=> 1.57x10”N.m -A recta:lgular loop has sides of length 0.06 m and 0.08 m. The wire carries a current of 10 A in the direction shown. What is the magnitude of the torque on the loop? Toque=N*1*B*A*sin8 To/que= (10Amp) (0.2T) (0.08x0.06) (sin60°)=8.31x10°N.m -A single circular loop of wire of radius 0.45 m carries a constant current of 2.4 A. The loop may be ..... to the direction of the magnetic field is 36°, the torque on the coil is 1.5 N.m. What is the magnitude of the uniform magnetic ﬂeld exerting this torq on the loop? To=N*I*B*A*sin6 => B=T. -A coil consists of 240 circular loops, each of radius 0.044 m, and carries a current of 2.2 A Determ:ne mag moment of the coil. Magnetic moment=N*I*A=> M=240loops*2.2Amps*n(0.044m)>=3.211 A.m? -Complete the foll statement: The mag field around a current-carrying, circular loop is most like that of: a short bar magnet. -A solenoid of length 0.250 m and radius 0.0250 m is comprised of 440 turns of wire. Determine the magn of the magnetic field at the center of the sole that carries a i of 12.0 A. B=po*N*I/L=41x10" *440x12Amp/0.25 =2.654x10%~2.21 x 10> T -An overhead electric power line carries a maximum current of 125 A. What is the magnitude of the maximum magnetic field at a point 4.50 m directly below the power line? B=pol/2nr=4nx107x125/21x4.5=5.56 x 10°T -Four long, straight wires are parallel to each other; and their cross-section forms a square. ...determine the mag of the total magn field at P, the center of the square. Square figs with contric circles B2 &B4 cancel each other, r=sqrt2a/2=sqr2x0.020/2=0.01414m, B=Hoi/2nr=4nx10"x8/21x0.01414m =1.1315 x 10~* T, total B from both wires at P will be 1.1315 x 10~ T times 2=2.3 x 10* T -The fig shows two concent metal loops, each carrying a current. The larger...What is the value of a current in the smaller Ioop that will result in zero total magnetic field at the center of the system? ¥ or ’ aomn- ] 0.10m —e 6OA" 1—Bg, ll[ 1—Igr;:8/0 06=1>/0.04=> 1,=8*0.04/0.06=5.3 A -Two long, straight, parallel wires separated by a distance d carry currents in opposite directions as shown in the fig .... The total magnetic field at point O is zero tesla. Arrows on top w/circles -(a) Determine the value of the current, |, in the top wire. |>=6Amp Distance from point C from both is d/2=0.10/2m, distance of point O from W, is d;=d+0.50d=1.5d, from w,, d,=0.5d, B at O is 0. Now, B1=B=> Hol1/27td1=pol2/27d,=>11/d1=1,/d>=> 0w 1;=6x1.5d/0.5d=18A. -(b) Determine the magnitude of the magnetic field at point Cif d = 0.10 m. Bc=B+B,=> Be=4nx107x (I,+1,)/2m(0.1/2)=9.6x10°T -Two long, straight wires separated by 0.10 m carry currents of 18 A and 6 A in the same direction as shown.w/circles (a) Determine the magnitude of the magnetic field at the point P. Vectors=arrows->Bp=B; + B;, B (due to a long wire)=p,(i)/2n(P distance), B due to both wires into page, add them. Bp=(uo(1,)/2n(r+d))+uo(1)/2n(r), B=4nx107x18/27(0.15)+4nx10"x 6/2n(0.05)=4.8 x 10° T -(b) What is the direction of the magnetic field at the point P? into the plane of the page -Each second, 1.25 x 10 electrons in a narrow beam pass through a small hole in a wall. The beam is perpendicular to the wall. Using Ampere's law, determine the magnitude of the magnetic field in the wall at a radius of 0.750 m from the center of the beam. I=Ne/t=1.25x10""x1.6x10""/1=2Amp, B=po(I)/27(r)=4nx107x2/27*0.75=2x10"x2/0.75=5.34 x 10" T -A long straight wire carries a 40.0 A current in the +x direction. At a particular instant, an electron moving at 1.0 x 10’ m/s in the +y direction is 0.10 m from the wire. The charge on the electron is 1.6 x 107*° C. What is the force on the electron at this - L. 1004 »oA 4O A O 1 wlectron O 10 an -l Wm-e instant? for B use Ampere law B=p,*I/2nr=8*10°T, |F|=q*v*B*sin(90)= -1.6*10°C*1.0*10’m/s*8*10*T *1=1.3 x 107 N in the +x direction -A wire is bent in shape of a circle of radius r = 0.10 m and carries a 20.0-A current in the shown directi CIRCLE ON TOP a) What is the direct of the magn field at the center of the loop? into the plane of the paper -(b) What is the magnitude of the magnetic field at the center of the loop?B=pu*1/2nr=4nx10"x 20/21*0.1=1.3 x 10* T -A long, straight wire carries a 10.0-A current in the +y direction as shown in the figure. Next to circle Next to the wire is a square copper loop that carries a 2.00-A current asshown.The length of each side of the square is 1.00 m. (a) What is the magnitude of the net magnetic force that acts on the loop?F=F-F>=poli[,L/2nd;-poli LL/2nd=> F=poliLL/2nt(1/d:-1/d> X . (b) What is the direction of the net magnetic force that acts on the Ioop? =X dlrection (c)Magnitude of induced emf if the 10A wire is cut and takes Ims to go to zero _ — 2 — -7, 2 2 3 Q! -A 1.2-kg rod that has a length of 1.0 m and a resistance of 5.0 (2 slides with constant speed down a pair of frictionless vertical field of magnitude 3.0 T is perpend to the plane formed by the rod and the rails as shown. Determi the speed of the rod. -A0. 60-1’ magnetic field is directed perpendlcular to the plane of a clrcular loop of radius 0.40 m. What is the magnitude of the magnetic flux through the loop? A=nr’, mag flux=¢=B*A*cos0=>0.60Txnx(0.40m)*(cos 0°=1)=0.30Wb -The Earth's magnetic field passes through a square tabletop with a magnitu of 4.95 x 10~ T and is directed at an angle of 165° relative to the normal of the tabletop If the tabletop has 1.50-m sndes, what is the magnitude of the magnetic flux through it?A X i X )8 -A circular copper loop is placed perpendlcular to a uniform magnetlc field of 0. 75 T. Determme the induced emf in the loop.since ¢=induc emf, ind emf=B*(rate change) =0.75*7.26*10°m?/s=5.4 x 103V -A conducting loop has an area of 0.065 m” and is positioned such that a uniform magnetic field is perpendicular to the plane of the loop. t0 0.30 T in 0.087 s, the average induced emf in the loop is 1.2 V. What is the initial value of the magnetic field? -A 250-turn solenoid carries a current of 9.0 A. The radius of the solenoid is 0.075 m; and its length is 0.14 m. Deter mag flux through the circular cross-sectional area at the center of the solenoid. N=(turns per length=250/0.14m=1785.714 turns), B=pu*N*L=47x107%1785.714*9 Amp=0.0201857T, A=mx(0.075)>-0.01766m?, $=B*A=0.0201857Tx0.01766m>=3.6 x 104 Wb -A magnetic field is directed perpendicular to the plane of a 0.15-m x 0.30-m rectangular coil consisting of 240 loops of wire. To induce an average emf of —2.5 Vin the coil, the mag field is increased from 0.1 T to 1.8 T det time interval $=A*B*N and e=-Ad/At , therefore, (-2.5V)=-(AB)*A*(N/At)=>At= (-AB)A*N/e= (-1.8T-.1T) (.15mx.30m) (240loops)/-2.5V=At=7.34s-The figure shows a uniform, 3.0-T magnetic field that is normal to the plane of a conducting, circular loop with a resistance of 1.5 Qand a radius of 0.024 m. The mag fie is direct out of the paper as shown. (a) What is the magnitude of the average induced emf in the loop if the magnitude of the magnetic field is doubled in 0.4 s? A=1*(0.024m)*=1.81x10*m?, AB=(Bf-B0)=3T, ¢=B*A, e=AxAB/At=(1.81x10"°m?)(3/0.45)=0.0136V=0.014V (b) What is the average current around the loop if the magnitude of the magnetic field is doubled in 0.4 s? 9.0 x 103 A, clockwise, i=e/R=0.0136V/1.50=9.05mA=9.0x10"*A (c) If the magnetic field is held constant at 3.0 T and the loop is pulled out of t what is the magnitude of the average induced emf in the loop?e=AxAB/At=3x(1.81x10"m/0.25)=0.027V=2.7 x 102V (d) If the magnetic field is held constant at 3.0 T and the loop is pulled out of the region that contain in enerqy dissipated in R? rate of energy Edissi=p=e%/R. p=(0.027)%/1.5=4.9x10"'W -A circuit is pulled with a 21-N force toward the right to maintain a constant speed v. At the instant shown, the loop is partially in and partially out of a uniform magnetic field that is directed into the paper. As the circuit moves, a 4.0-A current flows through a 6.0-C2 resistor. SHOWN on top Fig next to circle thingy 0.2 s, at what rate is

(a) Which one of the statements concerning this situation is true? The induced current flows clockwise around the circuit. Since the flux is decreasing. as per Lenz' law the induced emf will cause this change to cease, thus induced magnetic field will point into the page and induced current clockwise as per right hand rule. (b) What is the magnitude of v? 4.6 m/s power dissipated by the Resistor is given by P=i**R, F,y=i’R=>v=4’x6/21=4.6m/s -The angular speed of a motor is 262 rad/s. The back emf generated by the motor is 89.4 V. Assuming all other factors remain the same, determine the back emf if the angular speed of the motor is reduced to 131 rad/s. emf=NBAw, emf alpha w =emf is directly proportional to the ang velocity of the coil. hence, half of gmfi=em{’=89.4v/2=44.7v 11. A single conducting loop with an area of 2.0 m2 rotates in a uniform magnetic field so that the induced emf has a sinusoidal time dependence as shown. (a) With what angular frequency does the loop rotate? Time period for the loop=6.25-1.25=5s, angular freq=w=27/T=27/5s=1.256rad/s~1.26 rad/s (b) What is the period of the induced current? period for the loop=6.25-1.25=5s (c)Det the strength of the mag field in which the loop rotates. € mu=Baw=>B=€ma/A*w=7.5v/(2.0m?)(1.2566 rad/s)=2.984 T=3.0T -Determine the energy stored in a 7.09 x 107 H inductor that carries a 1.50-A current. inductance L.=7.09 x 10-7 H, Energy stored U=1/2Li*>=1/2x7.09 x 10-7 H*(1.5A)>= 7.98 x 107 ) -A solenoid with 1000 turns has a cross-sectional area of 7.0 cm? and length of 25 cm. How much energy is stored in the magnetic field of the solenoid when it carries a current of 10.0 A? L=N¢/i, $=BA, B=poNi/L. U=1/2p0 (NY/L)A#%, U=(1x4710- 7x1000x1000x7x10x10x100)/2x25x10000=0.1759 J=0.18 J -The figure shows a circular, conducting loop that is connected to a 5.0-V battery and a switch S. Immediately after the switch S is closed, the current through the loop changes at a rate of 15 A/s and the emf induced in the loop has a magnitude of 5.0 V. Determine the self-inductance of the coil. COIL next to waves L=emf/(di/dt). hence, L=5V/15a/s=0.333 H -A coil of wire with a resistance of 0.45 @ has a self-inductance of 0.083 H. If a 6.0-V battery is connected across the ends of the coil and the current in the circuit reaches an equilibrium value, what is the stored energy in the inductor? given equilib. Lo=V/R=6/0.45=13.33A., energy stored=1/2*Li’=0.5*0.083*13.33*13.33=7.37 J »7.4J -The current in the secondary coil of a step-up transformer is 1.25 A when the current in the primary coil is 0.30 A. Determine the turns ratio, Ny/N,, of the transformer. k=Ns/Np=ip/is, Np=primary turns, Ns=second. Ip=Primary I, Is=sec i, k=0.30A/1.25A=0.24 -A transformer changes 120 V across the primary to 1200 V across the secondary. If the secondary coil has 800 turns, how many turns does the primary coil have? 80 turns Np/Ns=Yp/Vs, Np=Ns(Vp/Vs)=(800)(120/1.200)=80 -A transf has 450 turns in its primary coil and 30 turns in its secondary coil. Which of the follow stgtem this transf is true? The power delivered to the secondary must be the same as that delivered to the primary. Ns=Vs*is=30*450, Np=V,*ip=30*450 -A small power plant produces a voltage of 6.0 kV and 150 A. The voltage is stepped up to 240 kV by a transformer before it is transmitted to a substation. The resistance of the transmission line between the power plant and the substation is 75 Q. (a) What is the current in the transmission line from the plant to the substation? Y,=6.0kV. I,=150A. Pi=V,*Ip. Pi=6x10°x150=9x10°W, Vs=240kV, VJ/Vp=Ip/Is =>(240kV/6kV)=150A/Is=>1s=3 . 75A~3.8A (b) What % of the power produced at the power plant is lost in transmission to the substation? 0.12 % Pi=i*R. PL=(3.8)"*75Q=1.083W. % power loss=(PL/P1)x100%=1.083w/9x10°w=0.12% -A loop is pulled with a force F to the right to maintain a constant speed of 8.0 m/s. The loop has a length of 0.15 m, a width of 0.080 m, and a resistance of 200.0 £2 At the instant shown, the loop is partially in and partially out of a uniform magnetic field that is directed into the paper. The magnitude of the field is 1.2 T. 34 GRAPHIC WITH OTHERS(a) What is the magnitude of the emf induced in the loop? 0.77 V., i=E/R, E=emf in V. R=resistance in Q. E= E=1.2Tx0.08mx8.0m/s=0.768V=0.77V -(b) What is the induced current in the loop? i=E/R=>i=0.768V/2009=3.84x10A -(c)_Determine the magnitude of the force required to pull the loop. F=iwB F=3.84x10°Ax0.08mx1.2T=3.7x10* N -A loop with a resistance of 4.0 Q is pushed to the left at a constant speed of 2.0 m/s by a 24 N force. At the instant shown in the figure, the loop is partially in and partially out of a uniform magnetic field. Aninduced current flows from left to right through the resistor. The length and width of the loop are 2.0 m and 1.0 m, respectively. 4t GRAPHIC WITH OTHERS -(a)_What is the direction of the magnetic field? into the paper -(b)_Determine the magnitude of the ind. i through the resistor,V=1*R, p=Fxy=24Nx2.0m/s=48W, p=i’*R=>i=sqrt p/R =sqrt48v/4.00=3 46A~3.5A. -(c) Determine the magnitude of the induced emf in the loop. €=i*R=3.5Ax40=14V -(d)_Determine the magnitude of the uniform magnetic field. e=Bly=>B=e/lv=14V/2x2=3.5T -(e)_At what rate is energy dissipated by the resistor?p=F*v=48W=~49 W -A flexible, circular conducting loop of radius 0.15 m and resistance 4.0 Q lies in a uniform magnetic field of 0.25 T. The loop is pulled on opposite sides by equal forces and stretched until its enclosed area is essentially zero m?, as suggested in the drawings. It takes 0.30 s to close the loop. -(a)_Determine the mag of the emf induced in the loop. €=-Ad/At=(nr’B)/At =(mx0.15mx0.15mx0.25T)/0.3s=5.9 x 102V -(b)_At what rate is heat generated in the loop? p=€*/R=(5.9x10V)*/4.00=8.7x10~*W -(c)Which one of the following phrases best describes the direction of the induced magnetic field generated by the current induced in the loop while the loop is being stretched? The induced current will flow clock-wise to produce an inward magnetic field to oppose the decreasing flux. into the page. --EM Waves c=3x10®m/s speed of light -_(Hz=m/s) - speed of light is 2.9979 x 10% m/s €.=8.854*10"'2 C%(N.m?), u=4n*107 -When a radio telescope observes a region of space between two stars, it detects electromagnetic radiation that has a wavelength of 0.21 m. This radiation was emitted by hydrogen atoms in the gas and dust located in that region. What is the frequency of this radiation? f=c/A, F=(3x10®m/s)/0.21m= 1.42 x 10° Hz -An FM radio station generates radio waves that have a frequency of 101.3 MHz, The frequency of the waves from a competing station have a frequency of 107.1 MHz, What is the difference in wavelength between the waves emitted from the two stations? f=c/A, A= ¢/f=>M= c/f1=(3x108m/s)/101.3MHz=2.9615m, A,= c/f,=(3x108m/s)/107.1MHz=2.8011m, A;-A; = 2.9615m-2.8011m=0.16 m — -A radio wave sent from the surface of the earth reflects from the surface of the moon and returns to the earth. The elapsed time between the generation of the wave and the detection of the reflected wave is 2.6444 s. Determine the distance from the surface of the earth to the surface of the moon. Note: The speed of light is 2.9979 x 108 m/s. time=2*dist/speed=>distanc=(speedxtime)/2= (2.9979 x 102 m/s x 2.6444 5)/2=3.9638 x 102 m - The average distance between the surface of the earth and the surface of the sun is 1.49 x 101 m. How much time, in minutes, does it take light leaving the surface of the sun to reach the earth? 8.3 min S=distance, c=speed of light t=s/c=>1.49 x 10! m/3 x 10®m/s= 496.6s/60=8.27min=~8.3 min -A cellular telephone transmits electromagnetic waves at a frequency of 835 MIHz, What is the wavelength of these waves? f=835MHz=>f=835 *106Hz, c=3 x 108m/s, A=Ff/c=>A=835 *10Hz /3 x 103m/s=0.359 m, -A distant space probe is programmed to emit a radio signal toward Earth at regular time intervals. One such pulse arrives on Earth 6.92 s after it is emitted from the probe. What is the approximate distance from the Earth to the probe? d=c*t=>d=3 x 10®m/s dividido 6.925=2.076*10°m=2.08*10°m -The amplitude of the electric field component of an electromagnetic wave is increased from E to 3E. What is the corresponding change in the intensity of the wave? 1=C€.E2 /2, E incre from E to 3E. 1=C€o(3E)? /2=> I=C€, (E?) /2=>I=9(CeE. (E?) /2), then, intensity increases by a factor of nine. -The peak value of the electric field component of an electromagnetic wave is E. At a particular instant, the intensity of the wave is of 0.020 W/m?2. If the electric field were increased to 5E, what would be the intensity of the wave? E1=E, i1=0.020 w/m?, E2=5E, 12=2, Intensity is proport to sguareofelecfieldampli, 11/12=E12/E22=>I,=1,*E22/E12=0.50 W/m? -An electromagnetic wave has an electric field with a peak value of 250 N/C. What is the average intensity of the wave? Eq=250n/c, c=3 x 108m/s, €0=8.854*10"'2 C?/(N.m?). I=CeoE,2 /2=> 1=(8.854*10"'2 C?/(N.m?)* 3 x 108m/s*(250N/C)2)/2=83 W/m? -An electromagnetic wave has an electric field with peak value 250.0 N/C. What is the average energy delivered to a surface with area 0.75 m2 by this wave in one hour? |=1/2*Ce&. E,2, power radiated by the wave P=I1*A, combine, P=(1/2*Ce.E.2)*A= P=(1/2*8.854*10"!2 C?/(N.m?)(3 x 108m/s) (250N/C)?)*0.75m2=62.23W, Power delivery in terms of energy P=E/t, to get energy delivered by the wave, E=P*t=>E=(62.23W)(1 hour)(3600/1hour)= 2.24 x 105J -An incandescent light bulb radiates uniformly in all directions with a total average power of 1.0 x 10> W. What is the maximum value of the magnetic field at a distance of 0.50 m from the bulb? P=1*10?W, r=0.5m, expression for | of light i=P/A where A=4nr?, i=P/4nr?, i=1*10>W /4n(0.5m)?=31.83w/m?, intensity of electromag waves is B>=2in,/c=> B=sqrt(2in./c)=>B=sqrt(2*31.83w/m?*4nx*107/3 x 10°m/s=5.16 x 10’ T=5.2 x 10’ T -A local radio station transmits radio waves uniformly in all directions with a total power of 7.50 x 10* W. What is the intensity of these waves when they reach a receiving antenna located 20.0 km from the transmitting antenna? i=P/A where A=4xr?, i=P/4nr’=> i=7.50*10°W/4n(20*10°m)*=1.49 x 10~° W/m? -A laser uniformly illuminates an area with green light that has an average intensity of 550 W/m?. What is the rms value of the electric field of this light? I=550 W/m?, €,=8.854*10"2 C?/(N.m?), c=3 x 10°m/s, Erms=sqrt((550w/m?)/ 8.854*10'> C*/(N.m*)*3 x 10°m/s)= 455 N/C -Electromagnetic waves are radiated uniformly in all directions from a source. The rms electric field of the waves is measured 35km from the source has an rms value of 0.012 N/C. Determine the average total power radi by the source. Erms=0.012N/C, r=35km=35000m, P=( §an,s_I"“A[|5!E>P=“0.012|z * 47(35*10°m)?) / 4w*107 *3 x 10°m/s=5.9 x 10° W -Unpolarized light of intensity S, passes through two sheets of polarizing material whose transmission axes make an angle of 60° with each other as shown in the figure. What is the intensity of the transmitted beam, $;? Intensi ssing through the first is reduced by a factor of %, on the second is S=(1 *cos60 *=(1/2)So(1/4)=S./8 -Linearly polarized light is incident of a sheet of polarizing material. The angle between the transmission axis and the incident electric field is 44°. What percentage of the incident intensity is transmitted? According to Malu’s law, the int of the polarized light is given as, izio*cos?(0), I/lo=_c0s?(44°)=0.5174, T%= 1/10*100%= 52 % -A linearly polarized beam of light is incident upon a group of three polarizing sheets which are arranged so that the transmission axis of each sheet is rotated by 45° with respect to the preceding sheet as shown. What fraction of the incident intensity is transmitted? GRAPHIC NEXT to star 1** polarizing is parallel, then i=1/2, second is i=1/2cos?45°=i=1/4 -Unpolarized light with an average intensity of 845 W/m? enters a polarizer with a vertical transmission axis. The transmitte: light then enters a second polarizer. The light that exits the second polarizer is found to have an average intensity of 225 W/m?. What is the orientation angle of the second polarizer relative to the first one? Given, avg i of a polarizer is 12=845/2 W/m?=422.5 W/m?, avg | of the 2nd polarizer is 1,=225 W/m?, form for avg i=1/2co0s?6°, where 0 is the orientation angle of the second i 20=225/422.5=0.5325=>c0s0=0.729=>0=cos* (0.729)=43.13° -Two polarizing sheets have their transmission axes parallel so that the intensity of unpolarized light transmitted through both of them is a maximum. Through what angle must either sheet be rotated if the transmitted intensity is 25 % of the incident intensity? I=1g9/2cos’0 0.25lo=19/2 * c0s?6, 0.25*2=cos*(6) => cosB@=sqrt0.50=0.7071, O=cos*(0.7071)=45° -A linearly polarized electromagnetic wave is sent through two sheets of polarizing material. The first sheet, A, is oriented so that its transmission axis makes an angle of 30° with respect to the incident electric field of the wave. The second sheet, B, is oriented so that its transmission axis makes an angle of 90° with the incident electric field of the wave. The incident beam has an electric field of peak magnitude E,; and average intensity So. Graphic with star SECOND (a) What is the peak value of the electric field amplitude after it goes through sheet A? I=lo*cos?(0), Et=sqgrt It=>Et=sqrt(lo*cos2(30°))=sqrt3/2= 0.87E, (b) What is the average intensity of the wave after it passes through A?S=So*cos?(0)=>It=l0*cos?(30°)=0.75S, (c) What is the average intensity of the wave after it passes through B?S=So*cos?(0)=>It=l0*cos?(90°)=0 S, (d) What is the average wave intensity after passing through both polarizing sheets? 0 21. The figure shows the time variation of the mag of the B of an electromagnetic wave produced by a wire antenna. (a) Determine the rms value of the electric field magnitude. Erms=E/sqrt2 =10/1.414=7.072 7.1 N/C (b) What is the peak value of the magnetic field?B=E/c=10/3*10°=3.3 x 10"* T (c) What is the rms value of the magnitude of the magnetic field? Brms=Erms/c=> Brms=(7.1 N/C)/ 3*10°= 2.366x10® T=2.4 x10° T(d) Determine the frequency of the wave. F=1/T=1/4x10°=2.5 x 10"® Hz(e) Determine the wavelength of the wave. A=c/f=3*10° /2.5 x 10° Hz=1.2 m (f) What is the avg total energy density of this electromagnetic wave?U=1/2* , *E?=1/2*8.85*10'2*10°=4.4 x 10*° J/m? (g) What is the average intensity of this electromagnetic wave? I=1/2*C* <, *E?=1/2*3*10°*8.85*10'2*10%=0.13 W/m? (h) What is the magnitude of the electric field at t = 3.34 x 107'° s? E=Emax*sin(w*t), where w=2x/T=1.57*10° rad/s, E=10sin{1.57*10° *3.34*10°)=5.0 N/C (i) What is the magnitude of the magnetic field at t = 6.67 x 107'° s? W=2xn/T=2x/4=n/2, E=10sin(xt/2*t)=> E=10sin(x/2*6.67 x 107 *° 5)=8.66N/C, B=E/c=>B=(8.66N/C)/ 3*10°=2.9 x 10T (i) What is the magnitude of the magnetic field at t = 6.0 x 107° s? At time 6.0 x 10° s the electric field E=0 as electromagnetic wave is touching the time axis E=c*B=>B=E/c=0/c=0 T zero teslal

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