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Q6.13 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 $2cm^{2}$ 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 $90^{0}$ 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 $7.5mC$ . The combined resistance of the coil and the galvanometer is $0.50\Omega$ . Estimate the field strength of magnet.

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best_answer

Given,

Area of search coil :

$A=2cm^2=2*10^{-4}m^2$

The resistance of coil and galvanometer

$R=0.5\Omega$

The number of turns in the coil:

$N=25$

Charge flowing in the coil

$Q=7.5mC=7.5*10^{-3}C$

Now.

Induced emf in the search coil

$e=N\frac{d\phi }{dt}=N\frac{\phi _{final}-\phi _{initial}}{dt}=N\frac{BA-0}{dt}=\frac{NBA}{dt}$

$e=iR=\frac{dQ}{dt}R=\frac{NBA}{dt}$

$B=\frac{RdQ}{NA}=\frac{0.5*7.5*10^{-3}}{25*2*10^{-4}}=0.75T$

Hence magnetic field strength for the magnet is 0.75T.

Posted by

Pankaj Sanodiya

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