A rod of length l slides against the perpendicular wall, such that the lowest point (A) of the rod slides with a speed v on horizontal surface as shown in figure. There exist a uniform magnetic field throughout the region. Now, what is the distance of point 'P' on the rod from end 'A' such that, the induced emf across 'P' and 'A' is maximum?
A rod of length l slides against the perpendicular wall, such that the lowest point (A) of the rod slides with a speed v on horizontal surface as shown in figure. There exist a uniform magnetic field throughout the region. Now, what is the distance of point 'P' on the rod from end 'A' such that, the induced emf across 'P' and 'A' is maximum?
Option 1 -
3l/8
Option 2 -
1/4
Option 3 -
3l/4
Option 4 -
1/2
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1 Answer
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Correct Option - 3
Detailed Solution:sin 60? = AP/ (lcos (30? )
AP = 3l/4
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