Class 12 Physics
CBSE and UP Board
Electromagnetic Induction
Chapter-6 Exercise 6.13
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 6 Question 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 2 cm² 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 coil and the galvanometer is 0.50 Ω . Estimate the field strength of magnet.
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Area of the small flat search coil, A = 2 cm2= 2 x 10-4 m2
Number of turns on the coil, N = 25
Total charge flowing in the coil, Q = 7.5 mC = 7.5 x 10-3 C
Total resistance of the coil and galvanometer, R = 0.50Ω
Induced current in the coil, I = [Induced emf (e)]/R————————Eq-1
Induced emf is given as: e = —N dφ/dt……………………Eq-2
Where, dφ = Charge in flux
Combining equations (1) and (2), we get
I = – (N dφ/dt)/R
=> I dt = -Ndφ/R —————————Eq-3
Initial flux through the coil,φi = BA
Where, B = Magnetic field strength
Final flux through the coil, φf = 0
Integrating equation (3) on both sides, we have
⌠Idt = -N/R Φi⌠Φf dΦ { NOTE- “⌠” is sign of integration}
But total charge ,Q = ⌠I dt.
Therefore, Q = -N/R (Φf-Φi) = -N/R (-Φi)= + NΦi/R
=> Q = NBA/R
Therefore,B = QR /NA
= (7.5 x 10-3 x 0.5)/(25 x 2 x 10-4)
= 0.75 T
Hence, the field strength of the magnet is 0.75 T.