First slide

Magnetic Flux

Question

A magnetising field of 1600 A/m produces a magnetic flux of 2.4 x $10^{- 5}$ Weber in an iron bar of cross section 0.2 ${cm}^{2}$ . Permeability of the bar is

Easy

A

$7.5 x 10^{- 3} T A^{- 1} m$
B

$7.5 x 10^{- 2} T A^{- 1} m$
C

$7.5 x 10^{- 4} {TA}^{- 1} m$
D

$75 x 10^{- 3} {TA}^{- 1} m$

Solution

$Φ = BA$
$B = \frac{Φ}{A} = \frac{2 .4 \times 10^{- 5}}{2 \times 10^{- 5}} = 1 .2$
$μ = \frac{B}{H} = \frac{1 .2}{1600} = \frac{12}{16000} = \frac{12}{16} \times 10^{- 3}$
$= 0.75 \times 10^{- 3}$
$= 7.5 \times 10^{- 4}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App