JEE MAIN - Physics (2013 (Offline) - No. 13)
Two short bar magnets of length $$1$$ $$cm$$ each have magnetic moments $$1.20$$ $$A{m^2}$$ and $$1.00$$ $$A{m^2}$$ respectively. They are placed on a horizontal table parallel to each other with their $$N$$ poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $$20.0$$ $$cm.$$ The value of the resultant horizontal magnetic induction at the mid-point $$O$$ of the line joining their centres is close to $$\left( \, \right.$$ Horizontal component of earth's magnetic induction is $$3.6 \times 10.5Wb/{m^2})$$
$$3.6 \times 10.5\,\,Wb/{m^2}$$
$$2.56 \times 10.4\,\,Wb/{m^2}$$
$$3.50 \times 10.4\,\,Wb/{m^2}$$
$$5.80 \times 10.4\,Wb/{m^2}$$
Explanation
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Given: $${M_1} = 1.20A{m^2}\,\,\,$$ and $$\,\,\,{M_2} = 1.00A{m^2}$$
$$r = {{20} \over 2}cm = 0.1m$$
$${B_{net}} = {B_1} + {B_2} + {B_H}$$
$${B_{net}} = {{{\mu _0}\left( {{M_1} + {M_2}} \right)} \over {{r^3}}} + {B_H}$$
$$ = {{{{10}^{ - 7}}\left( {1.2 + 1} \right)} \over {{{\left( {0.1} \right)}^3}}} + 3.6 \times {10^{ - 5}}$$
$$ = 2.56 \times {10^{ - 4}}\,\,wb/{m^2}$$
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