Question

a magnetising field of 2 into 10 raise to 3 ampere metre produces a magnetic flux density of 8π Tesla in an iron rod the relative permeability of the rod will be?​

Answer 1

Answer:  µr = 10⁴ H/m

Explanation:

Given data:

Magnetizing field of the iron rod, H = 2 * 10³ A/m

Magnetic flux density in an iron rod, B = 8π Tesla

To find: relative permeability of the iron rod

Let the relative permeability of the iron rod be “µr” and the permeability of vacuum or free space be “µo” is constant and is equal to 4π * 10⁻⁷ H/m.

The formula for the relative permeability, µr is given as

µr = (Magnetic permeability) / (Permeability of vacuum) = µ / µo

where, µ = (magnetic flux density) / (magnetising field) = B / H

∴ µr = B / (H * µo)

Or, µr = 8π / (2*10³ * 4π * 10⁻⁷)

Or, µr = 1/ (10⁻⁷ * 10³) = 1 / 10⁻⁴

Or, µr = 10⁻⁴ N/m

Hence, the relative permeability of the iron rod, µr is given as 10⁴ N/m.

Answer 2

Answer:

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