Question

1. An air-core solenoid 50 cm long has 4000 loops wound on it. Compute B in its interior when a current of 0.25 A exists in the winding.

Answer 1

Therefore the electric field in its interior is 2.512 × 10¯³ T.

Given : An air core solenoid 50 cm long has 4000 loops wound on it.

To find : The electric field, B in its interior when a current of 0.25A exists in the winding.

solution : number of turns per unit length, n = 4000/(0.5 m ) = 8000 turns/metre

current passing through solenoid, I = 0.25 A

using formula, B = $\mu_0ni$

= 4π × 10^-7 × 8000 × 0.25

= 4 × 3.14 × 8 × 0.25 × 10¯⁴ T

= 25.12 × 10¯⁴ T

= 2.512 × 10¯³ T

Therefore the electric field in its interior is 2.512 × 10¯³ T.

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