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If all linear dimensions of an inductor are tripled, then self inductance will become (keeping total number of turns per unit length constant)

1.  3 times

2.  9 times

3.  27 times

4. 13 times

L=un2LA = un2 L×πR2  Final length = 3L  Final radius of X-section = 3R  Final self inductance,      L'=un2×3L×π3R2          =27 un2LA      L'=27L

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The self inductance of a coil is L. Keeping the length and area same, the number of turns in the coil is increased to four times. The self inductance of the coil will now be 

(1) 14L

(2) L

(3) 4 L

(4) 16 L

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Pure inductance of 3.0 H is connected as shown below. The equivalent inductance of the circuit is

(1) 1 H

(2) 2 H

(3) 3 H

(4) 9 H

(1) The inductances are in parallel

Leq=L3=33=1H

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If a current of 10 A flows in one second through a coil, and the induced e.m.f. is 10 V, then the self-inductance of the coil is 

(1) 25H

(2) 45H

(3) 54H

(4) 1 H

(4) |e|=Ldidt10=L×101L=1H  

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The adjoining figure shows two bulbs B1 and B2 resistor R and an inductor L. When the switch S is turned off 

(1) Both B1 and B2 die out promptly

(2) Both B1 and B2 die out with some delay

(3) B1 dies out promptly but B2 with some delay

(4) B2 dies out promptly but B1 with some delay

(3) Current in B1 will promptly become zero while current in B2 will slowly tend to zero.

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In the figure magnetic energy stored in the coil is 

(1) Zero

(2) Infinite

(3) 25 joules

(4) None of the above

(3) i=VR=102=5A

U=12Li2=12×2×25=25J 

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