Two identical short bar magnets, each having | vedclass

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Question

(d)From figure ${B_{net}} = \sqrt {{B_a}^2 + {B_e}^2} $
$ = \sqrt {{{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{{2M}}{{{d^3}}}} \right)}^2} + {{\left(

Vedclass · Accepted Answer

$\sqrt 5 	imes {10^{ - 3}}$ $Tesla$

Vedclass · Answer

(d)From figure ${B_{net}} = \sqrt {{B_a}^2 + {B_e}^2} $
$ = \sqrt {{{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{{2M}}{{{d^3}}}} ight)}^2} + {{\left( {\frac{{{\mu _0}}}{{4\pi }}.\frac{M}{{{d^3}}}} ight)}^2}} $
$ = \sqrt 5 .\frac{{{\mu _0}}}{{4\pi }}.\frac{M}{{{d^3}}}$$ = \sqrt 5 	imes {10^{ - 7}} 	imes \frac{{10}}{{{{(0.1)}^3}}}$= $\sqrt 5 	imes {10^{ - 3}}$ $Tesla$.

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