The ratio of the radius of the two coils A and B is 1 : 2. The same magnet is first moved towards coil A and then towards coil B with the same speed from the same distance. If the number of turns is the same in both the coils then find the ratio of the induced in emf in coil A to coil B.

CONCEPT:

Faraday's first law of electromagnetic induction

• Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced.
• If the conductor circuit is closed, a current is induced which is called induced current.

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Faraday's second law of electromagnetic induction

• The induced emf in a coil is equal to the rate of change of flux linked with the coil.

\(\Rightarrow e=-N\frac{dϕ}{dt}\)

Where N = number of turns, dϕ = change in magnetic flux and e = induced e.m.f.

• The negative sign says that it opposes the change in magnetic flux which is explained by Lenz law.

CALCULATION:

Given \(\frac{R_A}{R_B}=\frac{1}{2}\), N_A = N_B = N, and \(\frac{dB_A}{dt}=\frac{dB_B}{dt}=B\)

∵ \(\frac{R_A}{R_B}=\frac{1}{2}\)

∴ \(\frac{A_A}{A_B}=\frac{\pi R_A^2}{\pi R_B^2}\)

\(\Rightarrow \frac{A_A}{A_B}=\frac{1}{4}\)

Where A_A = cross-sectional area of coil A and AB = cross-sectional area of coil B

• By the faradays second law, the emf induced in the coil A is given as,

\(\Rightarrow e_A=-N_A\frac{dϕ_A}{dt}\)

\(\Rightarrow e_A=-N_AA_A\frac{dB_A}{dt}\)

\(\Rightarrow e_A=-NA_AB\)      -----(1)

• By the faradays second law, the emf induced in the coil B is given as,

\(\Rightarrow e_B=-N_B\frac{dϕ_B}{dt}\)

\(\Rightarrow e_B=-N_BA_B\frac{dB_B}{dt}\)

\(\Rightarrow e_B=-NA_BB\)      -----(2)

By equation 1 and equation 2,

\(\Rightarrow \frac{e_A}{e_B}=\frac{-NA_AB}{-NA_BB}\)

\(\Rightarrow \frac{e_A}{e_B}=\frac{1}{4}\)

• Hence, option 1 is correct.