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NCERT Class XII Chapter
Electromagnetic Induction
Questions With Solutions

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Question : 13
Total: 17
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 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 = NBA = 25 B (2 × 10–4) = 50 × 10–4 × B Wb …(i)
Final flux through the coil is zero. Let coil is taken out in time ‘t’.
Magnitude of induced emf ε =
Δϕ
Δt
ε =
50×10−4B
t
 ... (ii)
Current in the coil
I =
ε
R
 =
50×10−4B
0.5t
 =
10−2B
t
 ... (iii)
Total charge flowing in the coil, q = It
q =
10−2B
t
 × t = 10−2 B or 7.5 × 10−3 = 10−2 B
So, magnetic field between poles, B = 0.75 T
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