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`        The magnetic field inside a long solenoid having 50 turns/cm is increased from 2.5 × 10^-3 T to 2.5 T when an iron core of cross-sectional area 4 cm is inserted into it. Find (a) the current in the solenoid, (b) the magnetization I of the core and (c) the pole strength developed in the core. `
3 years ago

396 Points
```										Sol. B base 1 = 2.5 x 10^-3   B base = 2.5
A = 4 x 10^-4 m^2 ,    n = 50 turns/cm 5000 turns/m
(a) B = μ base 0 ni,
⇒ 2.5 x 10^-3 = 4 π x 10^-1 x 5000 x i
⇒ i = 2.5x 10^-3/ 4π x 10^-7 x 5000 = 0.398 A ≈ 0.4 A
(b) I = B base 2 / μ base 0 – H = 2.5 / 4π x 10 ^ -7 – (B base 2 – B base 1) = 2.5 / 4π x 10^-7 – 2.497 = 1.99 x 10 ^ 6 ≈ 2 x 10^6
(c) I = M/V ⇒ ml/Al = m/A
⇒ m = IA = 2 x 10^6 x 4 x 10^-4 = 800 A-m

```
3 years ago
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