Magnetism Question 33
A cylindrical rod magnet has a length of 5 cm and a diameter of 1 cm. It has a uniform magnetisation of $5.30 × 10^3A/m$. What is its magnetic dipole moment
Options:
A) $ 1\times {{10}^{-2}}J/T $
B) $ 2.08\times {{10}^{-2}}J/T $
C) $ 3.08\times {{10}^{-2}}J/T $
D) $ 1.52\times {{10}^{-2}}J/T $
Show Answer
Answer:
Correct Answer: B
Solution:
Relation for dipole moment is, $ M=I\times A $ . Volume of the cylinder $ V=\pi r^{2}l $ , Where r is the radius and l is the length of the cylinder, then dipole moment, $ M=I\pi r^{2}l=(5.30\times 10^{3})\times \frac{22}{7}\times {{(0.5\times {{10}^{-2}})}^{2}}(5\times {{10}^{-2}}) $
$ =2.08\times {{10}^{-2}}J/T $
BETA
