Magnetism Question 191
Question: The dipole moment of a short bar magnet is 1.25 A-m2. The magnetic field on its axis at a distance of 0.5metre from the centre of the magnet is
Options:
A) $ 1.0\times {{10}^{-4}}Newton/amp-meter $
B) $ 4\times {{10}^{-2}}Newton/amp-metre $
C) $ 2\times {{10}^{-6}}Newton/amp-metre $
D) $ 6.64\times {{10}^{-8}}Newton/amp-metre $
Show Answer
Answer:
Correct Answer: C
Solution:
$ B=\frac{{\mu _{0}}}{4\pi }\frac{2M}{d^{3}}={{10}^{-7}}\times \frac{2\times 1.25}{{{( 0.5 )}^{3}}}=2\times {{10}^{-6}}N/A-m $
BETA
