An iron ring of mean circumference 30 cm and cross-section 1 $cm^2$ is wound uniformly with 300 turns of wire. When a current of 0.032 Amps flows in it, the flux produced is $2 × 10^-6Wb$. -

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 4M

Year: May 2013

Given:-

$Circumference; l = 30 cm= 0.3 m$

$A = 1 cm^2 = 10^-4 m^2$

Number of Turns,$ N = 300$

Current, $I = 0.032 A$

Flux,$ Φ = 2 × 10^-6Wb$

To Find:-

1) Flux Density (B)

2) Magnetic Field Intensity (H)

3) Permeability of Iron $(μ_I)$

Solution:-

Flux Density,

$$B= \frac{Φ}{A}$$

$$ B = \frac{2 × 10^{-6}}{10^{-4}}$$

$ ∴ B = 0.02 Wb/m^2$ …Ans

Magnetic Field Intensity,

$$ H =\frac{N.i}{l}$$

$$ ∴ H =\frac{300 × 0.032}{0.3}$$

$ ∴ H = 32 A-turns/m$ …Ans

Permeability of Iron,

$$ μ_I = \frac{B}{H}$$

$$μ_I = \frac{0.02}{32}$$

$μ_I = 6.25 × 10^{-4}m kg s^{-2} A^{-2}$