Q 13. We have a powerful loudspeaker magnet and have to measure the magnitude of the field between the poles of the speaker. And a small search coil is placed normal to the field direction and then quickly removed out of the field region, the coil is of the area and has 25 closely wound turns. Similarly, we can give the coil a quick  turn to bring its plane parallel to the field direction. We have measured the total charge flown in the coil by using a ballistic galvanometer and it comes to 7.5 mC. Total resistance after combining the coil and the galvanometer is . Estimate the field strength of the magnet.


 

Area of the small flat search coil, A=2cm2=2×104m2
Number of turns on the coil, N = 25
Total charge flowing in the coil, Q = 7.5mC=7.5×103C
Total resistance of the coil and galvanometer, R = 0.50 Ω
Induced current in the coil. I=Induced emf(e)R                             ...(1)
Induced emf is given as: e=Nddt                                 ...(2)
Where, dΦ = Charge in flux
Combining equations (1) and [2), we get

I=NdtRIdt=NR                                 ...(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=NRdϕ1
 But total charge, Q=IdtQ=NR(ϕjϕi)=NR(ϕi)=+iRQ=NBARB=QRNA=7.5×103×0.525×2×104=0.75T

Hence, the field strength of the magnet is 0.75 T.