PHYSICS ASSIGNMENT-9

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1 MPS/PHY-XII-11/A9 PHYSICS ASSIGNMENT-9 *********************************************************************************************************** 1. A wire kept long the north-south direction is llowed to fll freely. Will n emf be induced in the wire? No, becuse neither horizontl nor verticl component of erth s mgnetic field will be intercepted by the wire.. A trin is moving with uniform speed from north to south. (i) Will ny induced emf pper cross the ends of its le? (ii) Will the nswer be ffected if the trin moves from est to west? (i) Yes, emf will pper becuse the le is intercepting the verticl component of the erth s mgnetic field. (ii) No, here lso the le intercepts the verticl component of the erth s mgnetic field, so emf is induced cross the ends of the le. 3. As shown in the figure, conducting rod AB moves prllel to X-is in uniform mgnetic field, pointing in the positive Z-direction. The end A of the rod gets positively chrged. Is this sttement true? Give resons. Yes, the end A becomes positively chrged. According to right hnd thumb rule, the free electrons eperience mgnetic force in the direction from A to B. Deficit of electrons mkes end A positive while ecess of electrons mkes end B negtive. 4. A mgnet is moved in the direction indicted by n rrow between two coils AB nd CD s shown in the figure. Suggest the direction of current in ech coil. By Lenz s Lw, the ends of both the coils closer to the mgnet will behve s south poles. Hence the current induced in both the coils will flow clockwise when seen from the mgnet side.

2 5. In the figure, coil B is connected to low voltge bulb L nd plced prllel to nother coil A s shown. Eplin the following observtions: (i) Bulb lights (ii) bulb gets dimmer if the coil B is moved upwrds. (i) (ii) Bulb lights up due to the induced current set up in the coil B becuse of lternting current in coil A. Bulb gets dimmer when the coil b is moved upwrds becuse the flu linked with coil B decreses nd induced current lso decreses. 6. A coil A is connected to voltmeter V nd the other coil B to n.c source D. If lrge copper sheet C, is plced between the two coils, how does the induced emf in the coil A chnge due to current in the coil B? When the copper sheet is plced between the coils, eddy currents re induced in it which opposes the pssge of mgnetic flu. The rte of chnge of mgnetic flu linked with the coil A decreses. Hence the emf induced in coil A due to the chnge in current in coil B lso decreses. 7. A rod closing the circuit moves long U-shped wire t constnt speed v under the ction of the force F. The circuit is in uniform mgnetic field perpendiculr to its plne. Clculte F if the rte of genertion of het is P. The induced emf cross the ends of the rod is E = Blv. Current in the circuit I = E/ = Blv/. Mgnetic force on the conductor, F = ilb, towrds left. As there is no ccelertion, F =F => ilb = F or I = F/Bl As P = E I = Blv. F/Bl = Fv F = P/v

3 8. Two circulr coils, one of rdius r nd the other of rdius re plced coilly with their centers coinciding. For >r, obtin n epression for the mutul inductnce of the rrngement. Suppose current i flows through the outer circulr coil of rdius. The mgnetic field i t its center will be B = As the second co-illy plced coil hs very smll rdius r, the field B my be considered constnt over its cross-sectionl re. The flu linked with the smller coil will be i ϕ = BA = B π r = π r Therefore mutul inductnce M = ϕ π r = i 9. A copper rod of length L rottes with n ngulr speed ω in uniform mgnetic field B. Find the emf developed between the two ends of the rod. The field is perpendiculr to the motion of the rod. Suppose the rod completes one revolution in time T. Then chnge of flu = B X Are Swept = B πl where L is the length of the rod. Induced emf = Chnge in flu/time Or E = (B X πl )/T = BπL f. As f = ω/, therefore E =( BπL )( ω/) = ½ (BL ω) 1. ()A toroidl solenoid with n ir-core hs n verge rdius of 15cm, re of cross-section 1 cm nd 1 turns. Obtin the self-inductnce of the toroid. Ignore field vrition cross the cross-section of the toroid. (b) A second coil of 3 turns is wound closely on the toroid bove. If the current in the primry coil is incresed from zero to. A in.5 seconds, obtin the induced emf in the second coil. () The uniform mgnetic field set up inside solenoid is given by Ni B = ni = Totl flu linked with the N turns is Self inductnce of the toroid is ϕ N A 4π 1 (1) 1 1 L = = = i.15 = = H mh (b) Here N1 = 1, N = 3 dt=.5 s, di =. =. A Ni N ia ϕ = NBA = N.. A =

4 Therefore the emf induced in the second coil is di N N A di ε = = 1 M dt l dt π =.3V A squre loop of side 1 cm with its sides prllel to -is nd y-is moves with velocity of 8 cm/s in the positive -direction in n environment contining mgnetic field in the positive z-direction. The field is neither uniform in spce nor constnt in time. It hs grdient of 1-3 T/cm long the negtive -direction (i.e, it increses by 1-3 T/cm s one moves in the negtive -direction), nd it is decresing in time t the rte of 1-3 T/s. Determine the direction nd mgnitude of the induced current in the loop if its resistnce is 4.5 mω. Here A = =(.1m) = 144 X 1-4 m V=8cm/s=.8 m/s te of chnge of mgnetic field B with distnce is db/d = 1-3 T/cm = (1-3 T/1 - m) = 1-1 T/m te of chnge of mgnetic field B with time is db/dt = 1-3 T/s Induced emf due to chnge in field B with position is given by E1= dφ/dt = d(ba)/dt = A.(dB/dt)= A.(d/dt)(dB/d)= Av.(dB/d) = 144 X 1-4 X.8X 1-1 V = 115. X 1-6 V Induced emf due to chnge in field B with time t is E= dφ/dt = d(ba)/dt = A.(dB/dt) =144 X 1-4 X1-3 V = 14.4 X1-6 V Therefore Totl induced emf, E1+E = ( )X1-6 V =19.6X1-6 V As = 4.5mΩ=4.5 X 1-3 Ω Therefore induced current, i = E/ = (19.6X1-6 )/(4.5X1-3 ) A =.9 X 1 - A The two effects hve been dded up becuse both cuse decrese in flu in +ve z-direction. The direction of induced current is such s to increse the flu through the loop long the +ve z-direction. If for the observer the loop moves to the right, the current will be seen to be nticlockwise. 1. () Obtin n epression for the mutul inductnce between long stright wire nd squre loop of side s shown in the figure. (b) Now ssume tht the stright wire crries current of 5 A nd the loop is moved to the right with constnt velocity, v = 1m/s. Clculte the induced emf in the loop t the instnt when =.m. Tke =.1 m nd ssume tht the loop hs lrge resistnce.

5 () As shown in the figure, consider rectngulr strip of smll width dr of the loop t distnce r from the wire. Mgnetic field t the loction of the strip is i B = This field points normlly in to the plne of the loop. Are of the strip,a=dr. Mgnetic flu linked with the strip i dϕ = BA = dr Totl mgnetic flu linked with the squre loop, ϕ i r r= + = dϕ = dr = r= r= + 1 i i dr = [ln r] r= r i i + = [ln( + ) ln ] = ln = i ln(1 + ) (b) The squre loop is moving in non-uniform mgnetic field. The mgnetic flu linked with the loop t ny instnt is Induced emf set up in the loop, d i = v ln 1 d + i + i ϕ = ln(1 + ) d d d d ε = ϕ = ϕ = v ϕ dt d dt d = v 1 1+ Putting the vlues we get ε = 1.7 X1-5 V v = i π ( + ) 13. An ircrft with wing spn of 4 m flies with speed of 18 km/h in the estwrd direction t constnt ltitude in the northern hemisphere, where the verticl component of erth s mgnetic field is 1.75 X 1-5 T. Find the emf tht develops between the tips of the wings. The metllic prt between the wing-tips cn be treted s single conductor cutting flu-lines due to verticl component of erth s mgnetic field. So emf is induced between the tips of its wings. Here l = 4 m, B V = 1.75 X 1-5 T V= 18 km/h = 3m/s ε = B V lv =.1 V 14. The mgnetic flu through coil perpendiculr to the plne is vrying ccording to the reltion Φ = (5t 3 + 4t + t -5)Wb. Clculte the induced current through the coil t t = s, if the resistnce of the coil is 5Ω. ε = -dφ/dt t=s = 15.6 A

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