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It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm^(2) with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90^(@) turn to bring its plane parallel to the field direction. The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the soil and the galvanometer is 0.50 Omega. Estimate the field streng of magnet. |
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Answer» Solution :According to laws of ELECTROMAGNETIC induction, we know that `varepsilon=-(dphi_(B))/dt` Hence, induced current, `I=varepsilon/R=-N/R(dphi_(B))/dt` and the total induced charge `Q=int_"inital"^"final"dQ=int_(i)^(f)-N/R(dphi_(B))/dtdt` `=(-N)/Rint_(i)^(f)dphi_(B)=(-N)/R|phi_(B)|_(i)^(f)=-N/R(phi_(f)-phi_(i))` But `phi_(i ) = BA and phi_(f)=0`(because now plane of coil is parallel to the magnetic field) `Q=N/R(BA-0)=(NBA)/R or B = (QR)/(NA)` In present problem `N = 25,A=2cm^(2)=2XX10^(-4)m^(2),Q=7.5mC=7.5xx10^(-3)C and R=0.5Omeha` ` therefore B=(QR)/(NA)=((7.5xx10^(-3))(0.50))/(25xx2xx10^(-4))=0.75T.` |
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