Question

A solenoid is 1.5 m long and its inner diameter is 4.0 cm. It has three layers of windings of 1000 turns each and carries a current of 2.0 amperes. The magnetic flux for a cross-section of the solenoid is nearly(a) 2.5×10⁻⁷ weber
(b) 6.31×10⁻⁶ weber(c) 5.2×10⁻⁵ weber(d) 4.1×10⁻⁵ weber

Answer 1

length of solenoid, l = 1.5m

inner diameter of solenoid, d = 4cm = 0.04m

number of turns , N = 1000

current through solenoid, i = 2 Ampere.

we know, magnetic field across solenoid is given by, $B=\mu_0ni$

where $\mu_0$ is permeability of medium, n is number of turns per unit length and i is current through solenoid.

now, B = 4π × 10^-7 × (1000/1.5) × 2

= 4π × 10^-7 × 4000/3

= 16π/3 × 10^-4 Wb/m²

now, magnetic flux = BA

= 16π/3 × 10^-4 × π(0.04/2)²

= 16π/3 × 10^-4 × π × 0.02 × 0.02

= 6.31 × 10^-6 Wb

hence, option (b) is correct.