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QJEEM23Phff6e: JEE Main 2023 Physics - Class 12 JEEM23Phff6e Solution: A conducting circular loop of radius cm is placed perpendicular …

A conducting circular loop of radius 10π\frac{10}{\sqrt\pi} cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is :

Multiple Choice Options

  • A emf = 10 mV
  • B emf = 5 mV
  • C emf = 100 mV
  • D emf = 1 mV

Step-by-Step Verified Solution

amp;EMF=dϕdt=BA−0tamp;A=πr2=π(0.12π)=0.01amp; B=0.5amp;EMF=(0.5)(0.01)0.5=0.01 V=10 mV\begin{aligned} & \mathrm{EMF}=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{BA}-0}{\mathrm{t}} \\\\ & \mathrm{A}=\pi \mathrm{r}^2=\pi\left(\frac{0.1^2}{\pi}\right)=0.01 \\\\ & \mathrm{~B}=0.5 \\\\ & \mathrm{EMF}=\frac{(0.5)(0.01)}{0.5}=0.01 \mathrm{~V}=10~ \mathrm{mV}\end{aligned}

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