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7 Jan 2009 14:32:44 IST

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plz ans q no-44,45,50,52 of h c verma -2 pg no-122-123

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plz ans q no-44,45,50,52 of h c verma -2 pg no-122-123

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Prashant Mital

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7 Jan 2009 15:06:43 IST

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For Question 44

Considering that the loop is complete, we know that the electric field at the center will be zero. Now consider an element on the ring of length ∂l. The electric field exerted by this element at the loop center must be equal in magnitude and opposite in direction to the field exerted by the rest of the loop at the its center. So

E(at center due to the remaining wire) = E(at center due to ∂l)

Which gives the desired answer.

Vikram Saxena

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9 Jan 2009 04:14:19 IST

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52>>

this is not a SHM motion,but periodic

d = ut + 1/2 a t²

u = 0 and a = qe / m

t = (2md / qe)^1/2

total time taken to reach the wall is :

T = 2t

hence proved

Abhishek Srivastava

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24 Jan 2009 18:38:01 IST

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Abhishek Srivastava

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24 Jan 2009 18:38:39 IST

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Abhishek Srivastava

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24 Jan 2009 18:39:29 IST

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Abhishek Srivastava

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24 Jan 2009 18:40:14 IST

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Abhishek Srivastava

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24 Jan 2009 18:40:56 IST

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hey friend are you satisfied from all these answer please tell me is this mathod is beneficial for you

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