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 = µ_{0}ni ............(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

Concept: Magnetisation and Magnetic Intensity

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