JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 20)
A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20 $$ \times $$ 10–6 J/ T when a magnetic intensity of 60 $$ \times $$ 103 A/m is applied. Its magnetic susceptibility is
3.3 $$ \times $$ 10–4
2.3 $$ \times $$ 10–2
4.3 $$ \times $$ 10–2
3.3 $$ \times $$ 10–2
Explanation
x = $${1 \over H}$$
I = $${{Magnetic\,moment} \over {Volume}}$$
I = $${{20 \times {{10}^{ - 6}}} \over {{{10}^{ - 6}}}}$$ = 20 N/m2
x = $${{20} \over {60 \times {{10}^{ + 3}}}}$$ = $${1 \over 3} \times {10^{ - 3}}$$
= 0.33 $$ \times $$ 10$$-$$3 = 3.3 $$ \times $$ 10$$-$$4
I = $${{Magnetic\,moment} \over {Volume}}$$
I = $${{20 \times {{10}^{ - 6}}} \over {{{10}^{ - 6}}}}$$ = 20 N/m2
x = $${{20} \over {60 \times {{10}^{ + 3}}}}$$ = $${1 \over 3} \times {10^{ - 3}}$$
= 0.33 $$ \times $$ 10$$-$$3 = 3.3 $$ \times $$ 10$$-$$4
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