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If part of the magnet develops resistance and liquid helium boils away, rendering more and more of the magnet non super-conducting, how will this quench affect the time for the current to drop to half of its initial value? (a) The time will be shorter because the resistance will increase; (b) the time will be longer because the resistance will increase; (c) the time will be the same; (d) not enough information is given.

Short Answer

Expert verified

(a) The time will be shorter because the resistance will increase.

Step by step solution

01

Define R-L circuit

Some part of magnet develops resistance so magnet will act as R-L circuit. A circuit that includes both a resistor and an inductor, and possibly a source of electromotive force, is called an R-L circuit. The inductor helps to prevent rapid changes in current. Apply Kirchhoff鈥檚 loop rule in R-L circuit then on solving differential equation of current, current in R-L circuit is calculated.

02

Apply formula of current

Current in R-L circuit is given by equation,

i=R1-e-RLtWhere I is the current in circuit,is emf, R is resistance, L is inductance and t is time.

Solve above for t to get,

t=-LRlnii0t1R

Hence, magnet develops resistance which means the resistance increases and time will be shorter.

Therefore, the correct option is (a)The time will be shorter because the resistance will increase.

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