Question 17 of 153

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

A$\text{emf}=1mV$B$\text{emf}=10mV$C$\text{emf}=100mV$D$\text{emf}=5mV$

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Answer: (B) $\text{emf}=10mV$

Given: B=0.5T\text{at}t=0&B=0, at $t=0.5s$.
Magnetic field will decrease linearly as given below. We can say, value of magnetic field at $t=0.25\text{will be}B=0.25$.

Therefore,
${\epsilon }_{\text{ind}}=\left|\dfrac{\Delta \phi }{\Delta t}\right|=\left|\dfrac{\Delta \left(BA\right)}{\Delta t}\right|=A\left|\dfrac{\Delta \left(B\right)}{\Delta t}\right|$
$=\pi \times {\left(\dfrac{10}{\sqrt{\pi }}\right)}^{2}\times {10}^{-4}\times \left(\dfrac{0.25}{0.25}\right)$
$={10}^{-2}\times 1$
$=0.01V =10mV$