Question

an iron rod of 0.2 m long, 10 mm in diameter and of relative permeability 1000 is placed inside a long solenoid wound with 300 turns/meter. if a current of 0.5 A is passed through the rod, find the magnetic moment of the rod

Answer 1

Answer:

Magnetic moment of the rod is 2.35 Am

Explanation:

As we know that net magnetic field in the rod is given as

$B_{net} = B_o + B_{in}$

$B_{in} = B_{net} - \mu_o H$

$B_{in} = \mu_r\mu_o H - \mu_o H$

$B_{in} = (\mu_r - 1)\mu_o H$

$I = (\mu_r - 1) ni$

$I = (1000 - 1)(300)(0.5)$

$I = 149850$

now we know

$I = \frac{M}{V}$

so we have

$M = 149850 (\pi r^2) L$

$M = 149850(\pi \times 0.005^2)(0.2)$

$M = 2.35 Am$

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Topic : Magnetic Moment

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