A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced emf is
A. Six times per revolution
B. Once per revolution
C. Twice per revolution
D. Four times per revolution
Answer
581.4k+ views
Hint: When a wired loop is rotated in a magnetic field, then the change in emf is obtained from the differentiation of the magnetic flux. The direction of the electromotive force shall always remain the same. We will use Lenz's law to obtain the direction of induced electromotive force which is produced and hence find the frequency.
Complete answer:
Here, let us consider that the ring is starting to rotate from its initial position where it is perpendicular to the magnetic field. Thus, initially the maximum flux is passing through the loop. When the wired loop rotates, then the flux passing through the loop starts to decrease. When it becomes parallel to the magnetic field, then the flux becomes zero and then it starts to increase.
Here, the direction of the induced emf will always remain the same. Then as it rotates further, it would just keep on increasing, until it reaches maximum. According to Lenz law, the induced emf tends to oppose the flux when it is increasing and now since it has been decreasing, it will tend to increase it. Similarly when it reaches its initial position it changes the direction again and so there are two revolutions.
Hence option C is the correct answer.
Note: The flux of the magnetic field through the loop is given by the formula \[\phi = BA\cos \omega t\] and the differentiation of the magnetic flux gives emf which is given as \[\epsilon =\omega BA\sin \omega t\]. Here, the induced emf is the same, hence sine function equates to zero, hence \[\omega t = n\pi \] and thus, there are two revolutions for the frequency.
Complete answer:
Here, let us consider that the ring is starting to rotate from its initial position where it is perpendicular to the magnetic field. Thus, initially the maximum flux is passing through the loop. When the wired loop rotates, then the flux passing through the loop starts to decrease. When it becomes parallel to the magnetic field, then the flux becomes zero and then it starts to increase.
Here, the direction of the induced emf will always remain the same. Then as it rotates further, it would just keep on increasing, until it reaches maximum. According to Lenz law, the induced emf tends to oppose the flux when it is increasing and now since it has been decreasing, it will tend to increase it. Similarly when it reaches its initial position it changes the direction again and so there are two revolutions.
Hence option C is the correct answer.
Note: The flux of the magnetic field through the loop is given by the formula \[\phi = BA\cos \omega t\] and the differentiation of the magnetic flux gives emf which is given as \[\epsilon =\omega BA\sin \omega t\]. Here, the induced emf is the same, hence sine function equates to zero, hence \[\omega t = n\pi \] and thus, there are two revolutions for the frequency.
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