When a circular coil of radius \[1\,m\] and 100 turns is rotated in a horizontal uniform magnetic field, the peak value of emf Induced is \[100\,V\] . The coil is unwound and then rewound into a circular coil of radius \[2\,m\] . If it is rotated, now with the same speed, under similar conditions, the new peak value of emf developed is:
A. $50V$
B. $25V$
C. $100V$
D. $200V$
Answer
574.5k+ views
Hint:Let's get a sense of the peak value before we get into the question. The peak value of induced emf is the largest value of alternating emf. Now, in order to answer the question, we will use a ratio to compare the peak value in both instances, first when the radius was \[1m\] and then after the radius was doubled. If the emf peak values in the two scenarios are \[{V_1}\] and \[{V_2}\] , then we may use the ratio to determine the required answer.
Complete step by step answer:
A coil with \[N\] turns and area \[A\] rotates with angular velocity $\left( \omega \right)$ in a uniform magnetic field \[B\]. The peak value of the induced emf \[\left( {{V_{\max }}} \right)\] is therefore given by,
\[\left( {{V_{\max }}} \right){\text{ }} = {\text{ }}NAB\omega .\]
Since the radius has been doubled, the area of the coil in the second case is \[4A\]. However, because the length of wire required for a turn is doubled in the second scenario, the number of turns will be \[\dfrac{N}{2}\].If the peak values of the emf in the two situations are \[{V_1}\] and \[{V_2}\].We have;
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{100}}{{{V_2}}} = \dfrac{{NAB\omega }}{{\left[ {\left( {\dfrac{N}{2}} \right) \times \left( {4A} \right)B\omega } \right]}}$
$\Rightarrow \dfrac{{100}}{{{V_2}}} = \dfrac{1}{2}$
From which
$\therefore {V_2} = 200\,V$
Therefore, if it is rotated, now with the same speed, under similar conditions, the new peak value of emf developed is $200\,V$.
Therefore, the correct option is D.
Note:The difference between peak and rms value should not be misconstrued by students while solving problems. The maximum or highest value of current obtained in a single cycle is known as the peak value of current. The rms value of current, on the other hand, is defined as the root mean square value of current that may be determined when heat is dissipated in the resistor.
Complete step by step answer:
A coil with \[N\] turns and area \[A\] rotates with angular velocity $\left( \omega \right)$ in a uniform magnetic field \[B\]. The peak value of the induced emf \[\left( {{V_{\max }}} \right)\] is therefore given by,
\[\left( {{V_{\max }}} \right){\text{ }} = {\text{ }}NAB\omega .\]
Since the radius has been doubled, the area of the coil in the second case is \[4A\]. However, because the length of wire required for a turn is doubled in the second scenario, the number of turns will be \[\dfrac{N}{2}\].If the peak values of the emf in the two situations are \[{V_1}\] and \[{V_2}\].We have;
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{100}}{{{V_2}}} = \dfrac{{NAB\omega }}{{\left[ {\left( {\dfrac{N}{2}} \right) \times \left( {4A} \right)B\omega } \right]}}$
$\Rightarrow \dfrac{{100}}{{{V_2}}} = \dfrac{1}{2}$
From which
$\therefore {V_2} = 200\,V$
Therefore, if it is rotated, now with the same speed, under similar conditions, the new peak value of emf developed is $200\,V$.
Therefore, the correct option is D.
Note:The difference between peak and rms value should not be misconstrued by students while solving problems. The maximum or highest value of current obtained in a single cycle is known as the peak value of current. The rms value of current, on the other hand, is defined as the root mean square value of current that may be determined when heat is dissipated in the resistor.
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