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The Magnetic Field B Inside a Long Solenoid, Carrying a Current of 5.00 A, is 3.14 × 10−2 T. Find the Number of Turns per Unit Length of the Solenoid.

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प्रश्न

The magnetic field B inside a long solenoid, carrying a current of 5.00 A, is 3.14 × 10−2 T. Find the number of turns per unit length of the solenoid. 

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उत्तर

Given:
Magnitude of current, i = 5 A
Magnetic field intensity, B = 3.14 × 10−2 T
We know that the magnetic field inside a long solenoid having n turns per unit length is given by

`B = mu_0 ni`

`3.14 xx 10^-2 = 4 pi xx 10^-7 xx n xx5`

`⇒ n = (10^-2)/(20xx10^-7)`

`= 5 xx 10^3 = 5000 ` turns / m 

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पाठ 35: Magnetic Field due to a Current - Exercises [पृष्ठ २५२]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 35 Magnetic Field due to a Current
Exercises | Q 54 | पृष्ठ २५२

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