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NEET 2010
Exam
NEET
Year
2010
Subject
Physics
Chapter
Electromagnetic Induction

NEET 2010 · Physics · Electromagnetic Induction

A conducting circular loop is placed in a uniform magnetic field,...

A conducting circular loop is placed in a uniform magnetic field, B=0.025 TB=0.025 \mathrm{~T} with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1mms−11 \mathrm{mms}^{-1}. The induced emf when the radius is 2 cm2 \mathrm{~cm}, is

Options

  1. A2πμV2 \pi \mu \mathrm{V}
  2. BπμV\pi \mu \mathrm{V}
  3. Cπ2μV\frac{\pi}{2} \mu \mathrm{V}
  4. D2μV2 \mu \mathrm{V}

Correct answer

B. πμV\pi \mu \mathrm{V}

Step-by-step solution

Given: The magnetic field in the region is, (\mathrm{B}=0.025 \mathrm{~T}). Radius of the loop, (\mathrm{r}=2 \mathrm{~cm}=2 \times 10^{-2}) Constant rate at which radius of the loop shrimks, (\frac{\mathrm{dr}}{\mathrm{dt}}=1 \times 10^{-3} \mathrm{~ms}^{-1}) To find: The induced emf in the loop when radius is 2 cm. The magnetic flux linked with the loop is given by: (ϕ=BAcos⁡θ=B(πr2)cos⁡O∘⇒Bπr2\begin{aligned} & \phi=\mathrm{BA} \cos \theta \\ & =\mathrm{B}\left(\pi \mathrm{r}^2\right) \cos \mathrm{O}^{\circ} \\ & \Rightarrow \mathrm{B} \pi \mathrm{r}^2 \end{aligned}) The magnitude of induced emf is given as: (∣ϵ∣=dϕdt=ddt(Bπr2)=Bπ2rdrdt\begin{aligned} & |\epsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}} \\ & =\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B} \pi \mathrm{r}^2\right) \\ & =\mathrm{B} \pi 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \end{aligned}) Substitute the values. (∣ϵ∣=0.025×π×2×2×10−2×1×10−3=π×10−6 V⇒πμV\begin{aligned} & |\epsilon|=0.025 \times \pi \times 2 \times 2 \times 10^{-2} \times 1 \times 10^{-3} \\ & =\pi \times 10^{-6} \mathrm{~V} \\ & \Rightarrow \pi \mu \mathrm{V} \end{aligned})
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