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NCERT Class XII Chapter
Electromagnetic Induction
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Question : 13 of 17
Marks: +1, -0
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 cm2\mathrm{cm}^2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of field region. Equivalently, one can give it 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 coil and the galvanometer is 0.5 Ω. Estimate the field strength of magnet
Solution:  
Let the magnetic field between poles of loud speaker magnet is B.
Initial flux through the coil
ϕi\phi_i = NBA = 25 B (2 × 10−410^{-4}) = 50 × 10−410^{-4} × B Wb …(i)
Final flux through the coil is zero. Let coil is taken out in time ‘t’.
Magnitude of induced emf ε = ΔϕΔt\frac{\Delta\phi}{\Delta t}
ε = 50×10−4Bt\frac{50 \times 10^{-4} B}{t} ... (ii)
Current in the coil
I = εR\frac{\varepsilon}{R} = 50×10−4B0.5t\frac{50 \times 10^{-4} B}{0.5 t} = 10−2Bt\frac{10^{-2} B}{t} ... (iii)
Total charge flowing in the coil, q = It
q = 10−2Bt\frac{10^{-2} B}{t} × t = 10−210^{-2} B or 7.5 × 10−310^{-3} = 10−210^{-2} B
So, magnetic field between poles, B = 0.75 T
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