Question
Two identical short bar magnets, each having magnetic moment of $$10\,A{m^2},$$ are arranged such that their axial lines are perpendicular to each other and their centres be along the same straight line in a horizontal plane, If the distance between their centres is $$0.2\,m,$$ the resultant magnetic induction at a point midway between them is $$\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}H{m^{ - 1}}} \right)$$
A.
$$\sqrt 2 \times {10^{ - 7}}\,tesla$$
B.
$$\sqrt 5 \times {10^{ - 7}}\,tesla$$
C.
$$\sqrt 2 \times {10^{ - 3}}\,tesla$$
D.
$$\sqrt 5 \times {10^{ - 3}}\,tesla$$
Answer :
$$\sqrt 5 \times {10^{ - 3}}\,tesla$$
Solution :
From figure $${B_{{\text{net}}}} = \sqrt {B_a^2 + B_e^2} $$
$$ = \sqrt {{{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{{2M}}{{{d^3}}}} \right)}^2} + {{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{M}{{{d^3}}}} \right)}^2}} $$
$$ = \sqrt 5 .\frac{{{\mu _0}}}{{4\pi }}.\frac{M}{{{d^3}}} = \sqrt 5 \times {10^{ - 7}} \times \frac{{10}}{{{{\left( {0.1} \right)}^3}}} = \sqrt 5 \times {10^{ - 3}}\,tesla$$
From figure $${B_{{\text{net}}}} = \sqrt {B_a^2 + B_e^2} $$
$$ = \sqrt {{{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{{2M}}{{{d^3}}}} \right)}^2} + {{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{M}{{{d^3}}}} \right)}^2}} $$
$$ = \sqrt 5 .\frac{{{\mu _0}}}{{4\pi }}.\frac{M}{{{d^3}}} = \sqrt 5 \times {10^{ - 7}} \times \frac{{10}}{{{{\left( {0.1} \right)}^3}}} = \sqrt 5 \times {10^{ - 3}}\,tesla$$