The system containing the rails and the wire of the previous problem is kept vertically

Question:

The system containing the rails and the wire of the previous problem is kept vertically in a uniform horizontal magnetic field $\mathrm{B}$ that is perpendicular to the rails. It is found that the wire stays in equilibrium. If the wire ab is replaced by another wire of double its mass, how long will it take in falling through a distance equal to its length?

Solution:

$t=\sqrt{\frac{2 l}{g}}$

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Comments

Prince Gupta
Aug. 27, 2024, 6:35 a.m.
Thankyou sir aapka explanation aur mere teacher Saleem Ahmed sir fi ka explanation bhi same hai baaki YouTube aur other websites per explanation totally wrong hai
Anonymus
Aug. 13, 2024, 6:35 a.m.
In above problem when wire of double mass is placed then twice of mass*acceleration is equal to ma i.e 2ma = 2mg-mg due to which acceleration is equal to g/2 and time is equal 2(l/g)^1/2
None
Aug. 13, 2024, 6:35 a.m.
In above problem when wire of double mass is placed then twice of mass*acceleration is equal to ma i.e 2ma = 2mg-mg due to which acceleration is equal to g/2 and time is equal 2(l/g)^1/2
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