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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 $2 \mathrm{~cm}^2$ 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 quick $90^{\circ}$ turns 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 $7.5 \mathrm{mC}$. The combined resistance of coil and the galvanometer is $0.50 \Omega$. Estimate the field strength of magnet.
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Verified Answer
Let the magnetic field between poles of loud speaker magnet is $B$.
Initial flux through the coil $\phi_i=N B A=25 B\left(2 \times 10^{-4}\right)=50 \times 10^{-1} B$...(i)
Final flux through the coil is zero. Let coil is taken out in time ' $t$ '.
Magnitude of induced emf $e=\frac{\Delta \phi}{\Delta t}$ $e=\frac{50 \times 10^{-4} B}{t}$
Current in the coil $I=\frac{e}{R}=\frac{50 \times 10^{-4} B}{0.5 t}$
Total charge flowing in the coil $q=I t$ $q=\frac{10^{-2} B}{t} \times t=10^{-2} B$ or $7.510^{-3}=10^{-2} B$
So, magnetic field between poles, $B=0.75 \mathrm{~T}$
Initial flux through the coil $\phi_i=N B A=25 B\left(2 \times 10^{-4}\right)=50 \times 10^{-1} B$...(i)
Final flux through the coil is zero. Let coil is taken out in time ' $t$ '.
Magnitude of induced emf $e=\frac{\Delta \phi}{\Delta t}$ $e=\frac{50 \times 10^{-4} B}{t}$
Current in the coil $I=\frac{e}{R}=\frac{50 \times 10^{-4} B}{0.5 t}$
Total charge flowing in the coil $q=I t$ $q=\frac{10^{-2} B}{t} \times t=10^{-2} B$ or $7.510^{-3}=10^{-2} B$
So, magnetic field between poles, $B=0.75 \mathrm{~T}$
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