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5.Magnetism and Matter
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The dipole moment of a short bar magnet is $1.25\, A-m^2$. The magnetic field on its axis at a distance of $0.5$ metre from the centre of the magnet is
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
$6.64 \times {10^{ - 8}}\,Newton/amp - metre$
Solution
(c)$B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2M}}{{{d^3}}} = {10^{ – 7}} \times \frac{{2 \times 1.25}}{{{{\left( {0.5} \right)}^3}}} = 2 \times {10^{ – 6}}\,N/A – m$
Standard 12
Physics
