A small flat search coil of area 2 cm² with 25 closely wound turns, is positioned normal to the field direction, and then quickly rotated by 90°. The total charge flown in the coil is 7.5 mC. The resistance of the rod is 0.50 Ω. Estimate the field strength of magnetic field.

Solution

Given,
Area of coil, A = 2 cm²
Number of turns = 25
Total charge = 7.5 mC
Resistance of the rod = 0.50 Ω

The total charge flowing through the coil,

q = \int_{t_{1}}^{t_{2}} I dt = \frac{1}{R} \int_{t_{1}}^{t_{2}} E dt

where, induced EMF,

|E| = N \frac{dϕ}{dt}

\Rightarrow E dt = Ndϕ

Hence,

q = \frac{N}{R} \int_{ϕ_{1}}^{ϕ_{2}} dϕ = \frac{N}{R} (ϕ_{1} - ϕ_{2})

...(I)
Given,

N = 25, q = 7.5 \times 10^{- 3} C, ϕ_{2} = 0, R = 0.50 Ω

Putting these values in equations (i), we get

ϕ_{1} = 1.5 \times 10^{- 4} Wb, A = 2.0 \times 10^{- 4} m^{2}

Q = \frac{qR}{N}

(From I)

∴

B = \frac{ϕ_{1} - ϕ_{2}}{A} = 0.75 T .

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