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Q.

A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of $2 A$ . If the number of turns is 1000 per metre, magnetisation of core is approximately given by: -

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a

$4 \times 10^{5} A / m$

b

$9 \times 10^{5} A / m$

c

$6 \times 10^{5} A / m$

d

$8 \times 10^{5} A / m$

answer is C.

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Detailed Solution

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The magnetic intensity H is $H = nI = 1000 \times 2.0 = 2 \times 10^{3} A / m.$
The magnetic field B is given by
$B = μ_{r} μ_{0} H = 400 \times 4 π \times 10^{- 7} (N / A^{2}) \times 2 \times 10^{3} (A / m) = 1.0 T$
Magnetisation is given by $M = (B - μ_{0} H) / μ_{0} \approx 8 \times 10^{5} A / m$

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