An iron rod of volume 10–3 m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :
Solution
Given, V = 10<sup>–3</sup> m<sup>3</sup>
= Al
<br><br>I = 0.5A
<br><br>$\mu$<sub>r</sub> = 1000
<br><br>n = 10 turns/cm = ${{10} \over {{{10}^{ - 2}}}}$ turn/m = 1000 turn/m
<br><br>Magnetic moment, M = NIA($\mu$<sub>r</sub> - 1)
<br><br>= (nl)IA($\mu$<sub>r</sub> - 1)
<br><br>= nI(Al)($\mu$<sub>r</sub> - 1)
<br><br>= 1000 × 0.5 × 10<sup>–3</sup> (1000 – 1)
<br><br>= 0.5 × (999) = 499.5
<br><br>$\simeq$ 500 = 5 $\times$ 10<sup>2</sup> Am<sup>2</sup>
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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