NEET-XII-Physics
37: Magnetic Properties of Matter
- #1The 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.Ans : 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`` \left(B\right)`` 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`` \left(B\right)`` and magnetic intensity`` \left(H\right)`` is given by
H = `` \frac{B}{{\,\mathrm{\,\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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