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A rectangular, a square, a circular and an elliptical loop, all in the (x–y) plane, are moving out of a uniform magnetic field with a constant velocity, $\vec{V} = v\hat{i}$. The magnetic field is directed along the negative z-axis direction. The induced emf, during the passage of these loops, coming out of the field region, will not remain constant for
A
Any of the four loops
B
The rectangular, circular and elliptical loops
C
The circular and the elliptical loops
D
Only the elliptical loop
Detailed Solution
Induced emf $= \frac{d\phi}{dt} = B\frac{dA}{dt}$, where A is the area of the loop still inside the field.
As a loop moves out with velocity v, $\frac{dA}{dt} = lv$, where l is the length of the loop along the boundary of the field region at that instant; so emf = Blv.
For the rectangular and square loops, the length l cut by the boundary stays the same throughout, so the emf remains constant.
For the circular and elliptical loops, the chord cut by the boundary keeps changing as the loop comes out, so $\frac{dA}{dt}$ and hence the emf change with time.
Hence the induced emf will not remain constant for the circular and the elliptical loops.
As a loop moves out with velocity v, $\frac{dA}{dt} = lv$, where l is the length of the loop along the boundary of the field region at that instant; so emf = Blv.
For the rectangular and square loops, the length l cut by the boundary stays the same throughout, so the emf remains constant.
For the circular and elliptical loops, the chord cut by the boundary keeps changing as the loop comes out, so $\frac{dA}{dt}$ and hence the emf change with time.
Hence the induced emf will not remain constant for the circular and the elliptical loops.