Problem 4. A circular wire loop 40 cmin diameter has 100-!resistance and lies in a horizontal plane. A uniform magnetic field points vertically downward, and in 25 msit increases linearly from 5.0 mT to 55 mT.Find the magnetic flux through the loop at (a) the beginning and (b) the end of the 25 msperiod. (c) What is the loop current during this time? (d) Which way does this current flow? Solution (a) As in the previous solution, !B=B"A=14#d2B=14#(40 cm)2(5 mT)=6.28$10%4Wb at t1=0, and(b) 6.91!10"3Wb at t2=25 ms.(c) Since the field increases linearly, d!B=dt="!B="t=(6.91#0.628)$10!3Wb=25 ms=0.251 V.From Faraday’s law, this is equal to the magnitude of the induced emf, which causes a current I=E=R=0.251 V=100 !=2.51 mAin the loop. (d) The direction must oppose the increase of the external field downward, hence the induced field is upward and Iis CCW when viewed from above the loop. Problem 6. A conducting loop of area Aand resistance Rlies at right angles to a spatially uniform magnetic field. At time t=0the magnetic field and loop current are both zero. Subsequently, the current increases according to I=bt2,where bis a constant with the units A/s2.Find an expression for the magnetic field strength as a function of time. Solution The induced current and the derivative of the magnetic field strength are related as in the previous problem, dB=dt=IR=A=(bR=A)t2.Integration yields B(t)=(bR=A)t3=3,where B(0)=0was specified. Problem 9. A square wire loop of side land resistance Ris pulled with constant speed vfrom a region of no magnetic field until it is fully inside a region of constant, uniform magnetic field Bperpendicular to the loop plane. The boundary of the field region is parallel to one side of the loop. Find an expression for the total work done by the agent pulling the loop. Solution The loop can be treated analogously to the situation analyzed in Section 31-3, under the heading “Motional EMF and Lenz’s Law”; instead of exiting, the loop is entering the field region at constant velocity. All quantities have the same magnitudes, except the current in the loop is CCW instead of CW, as in Fig. 31-13. Since the applied force acts over a displacement equal to the side-length of the loop, the work done can be calculated directly: Wapp=Fapp!l=(IlB)l=Il2B.But, I=E=R=d!B=dt=R=d=dt(Blx)=R=Blv=R,as before, so Wapp=B2l3v=R.[Alternatively, the work can be calculated from the conservation of energy: I=Blv=R,Pdiss=I2R=(Blv)2=R,and Wapp=Pdisst=[(Blv)2=R](l=v).]Problem 14. A square wire loop 3.0 mon a side is perpendicular to a uniform magnetic field of 2.0 T.A 6-V light bulb is in series with the loop, as shown in Fig. 31-45. The magnetic field is reduced steadily to zero over a time !t.(a) Find !tsuch that the light will shine at full brightness during this time. (b) Which way will the loop current flow?
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