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5.Magnetism and Matter
easy
A small bar magnet has a magnetic moment $1.2 \,A-m^2$. The magnetic field at a distance $ 0.1\, m $ on its axis will be : ($\mu_0 = 4\pi \times 10^{-7} \,T-m/A$)
A
$1.2 \times 10^{-4} \,T$
B
$2.4 \times 10^{-4} \,T$
C
$2.4 \times 10^{4} \,T$
D
$1.2 \times 10^{4}\, T$
Solution
(b)$B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2M}}{{{d^3}}} \Rightarrow B = {10^{ – 7}} \times \frac{{2 \times 1.2}}{{{{\left( {0.1} \right)}^3}}} = 2.4 \times {10^{ – 4}}\,T$
Standard 12
Physics
