JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 9)
An iron rod of volume 10–3 m3 and relative
permeability 1000 is placed as core in a
solenoid with 10 turns/cm. If a current of 0.5 A
is passed through the solenoid, then the
magnetic moment of the rod will be :
5 $$ \times $$ 102 Am2
0.5 $$ \times $$ 102 Am2
500 $$ \times $$ 102 Am2
50 $$ \times $$ 102 Am2
Explanation
Given, V = 10–3 m3
= Al
I = 0.5A
$$\mu $$r = 1000
n = 10 turns/cm = $${{10} \over {{{10}^{ - 2}}}}$$ turn/m = 1000 turn/m
Magnetic moment, M = NIA($$\mu $$r - 1)
= (nl)IA($$\mu $$r - 1)
= nI(Al)($$\mu $$r - 1)
= 1000 × 0.5 × 10–3 (1000 – 1)
= 0.5 × (999) = 499.5
$$ \simeq $$ 500 = 5 $$ \times $$ 102 Am2
I = 0.5A
$$\mu $$r = 1000
n = 10 turns/cm = $${{10} \over {{{10}^{ - 2}}}}$$ turn/m = 1000 turn/m
Magnetic moment, M = NIA($$\mu $$r - 1)
= (nl)IA($$\mu $$r - 1)
= nI(Al)($$\mu $$r - 1)
= 1000 × 0.5 × 10–3 (1000 – 1)
= 0.5 × (999) = 499.5
$$ \simeq $$ 500 = 5 $$ \times $$ 102 Am2
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