Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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The Magnetic Intensity H at the Centre of a Long Solenoid Carrying a Current of 2.0 A, is Found to Be 1500 a M−1. Find the Number of Turns per Centimetre of the Solenoid. - Physics

Sum

The magnetic intensity H at the centre of a long solenoid carrying a current of 2.0 A, is found to be 1500 A m−1. Find the number of turns per centimetre of the solenoid.

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Solution

Here,

Current in the solenoid, I = 2 A

Magnetic intensity at the centre of long solenoid, H = 1500 Am−1

Magnetic field produced by a solenoid (B) is given by,

B = µ0ni  ............(1)

Here, n = number of turns per unit length

i = electric current through the solenoid

Also, the relation between magnetic field strength (B) and magnetic intensity (H) is given by,

\[H=\frac{B}{\mu_0}.............(2)\]

From equations (1) and (2), we get:-

H = ni

⇒ 1500 A/m = n × 2

⇒ n = 750 turns/meter

⇒ n = 7.5 turns/cm

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 15 Magnetic Properties of Matter
Q 1 | Page 286
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