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(A) According to Faraday's law of electromagnetic induction, the induced emf $\varepsilon$ is given by $\varepsilon = N \left| \frac{\Delta \phi}{\Del

Vedclass · Accepted Answer

$70$

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(A) According to Faraday's law of electromagnetic induction, the induced emf $\varepsilon$ is given by $\varepsilon = N \left| \frac{\Delta \phi}{\Delta t} ight|$.Here, $\Delta \phi = (\Delta B) A$, where $A = \pi r^2$ is the area of the coil.Given: $B_i = 5000 \,T$, $B_f = 3000 \,T$, $\Delta t = 2 \,s$, $\varepsilon = 22 \,V$, and diameter $d = 0.02 \,m$.The radius $r = \frac{d}{2} = 0.01 \,m$.The change in magnetic field $\Delta B = |B_f - B_i| = |3000 - 5000| = 2000 \,T$.The area $A = \pi (0.01)^2 = 0.0001 \pi \,m^2$.The change in flux $\Delta \phi = (\Delta B) A = 2000 	imes 0.0001 \pi = 0.2 \pi \,Wb$.Substituting these values into the emf formula: $22 = N \left( \frac{0.2 \pi}{2} ight)$.$22 = N (0.1 \pi)$.Using $\pi \approx 3.14$, $22 = N (0.314) \Rightarrow N = \frac{22}{0.314} \approx 70$.Thus, the number of turns is $70$.

$A$ coil is placed perpendicular to a magnetic field of $5000 \,T$. When the field is changed to $3000 \,T$ in $2 \,s$, an induced emf of $22 \,V$ is produced in the coil. If the diameter of the coil is $0.02 \,m$, then the number of turns in the coil is:

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