Question

A closely wound coil of 100 turns and area of crosssection 1 cm² has a coefficient of self-induction 1 mH. The magnetic induction in the centre of the core of the coil when a current of 2A flows in it, will be
(a) 0.022 Wbm⁻²(b) 0.4 Wb m⁻²(c) 0.8 Wb m⁻²(d) 1 Wb m⁻²

Answer 1

CORRECT ANSWER IS : option a = 0.022 wb/m2

Explanation :

A= 1cm²= 1x 10⁻² m²

I=2A

N=100

B=μ₀NI/2r

=4π×10⁻⁷×100×2×π/2×10⁻²

=0.022 wb/m²

Answer 2

Answer:

a) 0.022 Wbm⁻²

Explanation:

Length of the coil = 100 (Given)

Cross section of the coil = 1cm² (Given)

Coefficient of self-induction = 1 mH (Given)

Current in the coil = 2A (Given)

Cross section = A= 1cm²= 1x 10⁻² m²

According to the formula used to calculate the magnetic field at the center of a thin coil -

B=μ₀NI/2r

where N is the number of turns  and r is the radius of center of loop current

= 4π×10⁻⁷×100×2×π/2×10⁻²

= 0.022

Therefore, the magnetic induction in the centre of the core of the coil when a current of 2A flows in it, will be 0.022 Wbm⁻².