Question

It is desired to measure the magnitude of field between the poles of apowerful loud speaker magnet. A small flat search coil of area 2 cm^2with 25 closely wound turns, is positioned normal to the fielddirection, and then quickly snatched out of the field region.Equivalently, one can give it a quick 90° turn to bring its planeparallel to the field direction). The total charge flown in the coil(measured by a ballistic galvanometer connected to coil) is7.5 mC. The combined resistance of the coil and the galvanometer is0.50 W. Estimate the field strength of magnet.

Answer 1

The given area of the small flat coil is 2  cm 2 .

A=2× 10 −4   m 2

Number of turns, N=25.

Total charge flown in the coil, Q=7.5 mC=7.5× 10 −3  C.

The combined resistance of the coil and galvanometer, R=0.50 Ω.

The formula to calculate the induced emf is,

I= Induced emf Resistence = e R (1)

The formula to calculate the induced emf is,

e=−N dϕ dt (2)

Where, ϕ is the flux linked to the coil.

From equation (1) and (2), the formula for induced current becomes,

I= −N dϕ dt R I=− N R dϕ dt Idt=− N R dϕ (3)

Initial flux through the coil is given as ϕ i =BA.

Here, B is the magnetic field strength and A is the area of the coil.

Integrate both sides of equation (3).

∫ Idt =− N R ∫ ϕ i ϕ f dϕ Q=− N R [ ϕ ] ϕ i ϕ f [ ∵I= dQ dt ] =− N R [ ϕ f − ϕ i ]

Now ϕ i =BA and ϕ f =0.

Q= NBA R B= QR NA

Substitute the value in the above equation.

B= 7.5× 10 −3 ×0.5 25×2× 10 −4 = 3.75× 10 −3 50× 10 −4 =0.75 T

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