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A square loop of side $2 \mathrm{~cm}$ enters a magnetic field with a constant speed of $2 \mathrm{~cm} \mathrm{~s}^{-1}$ as shown. The front edge enters the field at $t=0 \mathrm{~s}$. Which of the following graph correctly depicts the induced emf in the loop?
(Take clockwise direction positive)
Options:
(Take clockwise direction positive)
- A
- B
- C
- D
Solution:
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Verified Answer
The correct answer is:
As the loop moves, there are three cases.
(i) When the loop moves fully outside the magnetic field, there is no change in flux.
$\therefore \varepsilon=\frac{d \phi}{d t}=0$
(ii) When the loop is entering/leaving the area of magnetic field.
$\begin{gathered}\varepsilon=B L V=(0.9)\left(2 \times 10^{-2}\right)\left(2 \times 10^{-2}\right) \\ =2 \times 10^{-4} \mathrm{~V}\end{gathered}$
As flux is in opposite direction to magnetic field.
$\varepsilon$ when loop enters the field $=-2 \times 10^{-4} \mathrm{~V}$
$=2 \times 10^{-4} \mathrm{~V}$
(iii) When moving inside the magnetic field.
$\frac{d \phi}{d t}=0$
$\therefore$ No emf is induced.
(i) When the loop moves fully outside the magnetic field, there is no change in flux.
$\therefore \varepsilon=\frac{d \phi}{d t}=0$
(ii) When the loop is entering/leaving the area of magnetic field.
$\begin{gathered}\varepsilon=B L V=(0.9)\left(2 \times 10^{-2}\right)\left(2 \times 10^{-2}\right) \\ =2 \times 10^{-4} \mathrm{~V}\end{gathered}$
As flux is in opposite direction to magnetic field.
$\varepsilon$ when loop enters the field $=-2 \times 10^{-4} \mathrm{~V}$
$=2 \times 10^{-4} \mathrm{~V}$
(iii) When moving inside the magnetic field.
$\frac{d \phi}{d t}=0$
$\therefore$ No emf is induced.
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