JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 3)
A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. The corecivity of the bar magnet is :
285 A/m
2600 A/m
520 A/m
1200 A/m
Explanation
Coercivity, H = $${B \over {{\mu _0}}}$$
Inside solenoid the magnetic field,
B = $$\mu $$0ni
$$ \therefore $$ H = $${{{\mu _0}ni} \over {{\mu _0}}}$$
= ni
= $${N \over \ell } \times i$$
= $${{100} \over {0.2}} \times 5.2$$
= 2600 A/m
Inside solenoid the magnetic field,
B = $$\mu $$0ni
$$ \therefore $$ H = $${{{\mu _0}ni} \over {{\mu _0}}}$$
= ni
= $${N \over \ell } \times i$$
= $${{100} \over {0.2}} \times 5.2$$
= 2600 A/m
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