Search any question & find its solution
Question:
Answered & Verified by Expert
A bar magnet having length $5 \mathrm{~cm}$ and area of cross-section $4 \mathrm{~cm}^{2}$ has magnetic
moment $2 \mathrm{Am}^{2}$. If magnetic susceptibility is $5 \times 10^{-6}$, the magnetic intensity will
be
Options:
moment $2 \mathrm{Am}^{2}$. If magnetic susceptibility is $5 \times 10^{-6}$, the magnetic intensity will
be
Solution:
1126 Upvotes
Verified Answer
The correct answer is:
$2 \times 10^{10} \frac{\mathrm{A}}{\mathrm{m}}$
$\begin{aligned} \text { Volume of the magnet } \mathrm{V} &=\ell \times \mathrm{A} \\ &=5 \mathrm{~cm} \times 4 \mathrm{~cm}^{2}=20 \mathrm{~cm}^{3} \\ &=20 \times 10^{-6} \mathrm{~m}^{3}=2 \times 10^{-5} \mathrm{~m}^{3} \\ \text { magnetic moment }=2 & \mathrm{Am}^{2} \\ \therefore \text { Magnetization } \mathrm{M} &=\frac{2 \times 10^{-5}}{2}=10^{-5} \mathrm{~A} / \mathrm{m} \\ \text { Magnetic intensity } \mathrm{H} &=\frac{\mathrm{M}}{\chi}=\frac{10^{-5}}{5 \times 10^{-6}}=2 \times 10^{10} \mathrm{~A} / \mathrm{m} \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.