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प्रश्न
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.
उत्तर
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
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