Magnetic Properties of Matter

JEE Main level, numerical answer, medium

13 of 57

Question 13 of 57

The current required to be passed through a solenoid of 15cm15cm length and 6060 turns in order to demagnetise a bar magnet of magnetic intensity 2.4×103Am–12.4\times {10}^{3}A{m}^{–1} is ________ AA.

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Answer: 6

The magnetic intensity is
H=Bμ0−MH=\dfrac{B}{{\mu }_{0}}-M
For M=0M=0, it can be written
H=(Bμ0)=niH=\left(\dfrac{B}{{\mu }_{0}}\right)=ni
The data given is
H=2.4×103Am−1l=15×10−2mN=60H=2.4\times {10}^{3}A{m}^{-1} l=15\times {10}^{-2}m N=60
Using the relation, n=Nln=\dfrac{N}{l} and H=niH=ni, the value of the current is,
i=Hn=2.4×103×15×10−260=6Ai=\dfrac{H}{n}=\dfrac{2.4\times {10}^{3}\times 15\times {10}^{-2}}{60}=6A

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