A) halved
B) doubted
C) \[{{l}_{2}}\] times the original value
D) unaffected
Correct Answer: B
Solution :
The magnetic field inside the solenoid\[=\frac{v'}{v}=\frac{v/2}{v}=\frac{1}{2}=0.5\] Flux, \[\phi =BnlA={{\mu }_{0}}{{n}^{2}}lAi\]where, A = area of cross-section. As, \[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CON}{{\text{H}}_{\text{2}}}\] Here, n = number of turns per unit length \[\text{C}{{\text{H}}_{\text{3}}}NCO\] When N and \[\text{C}{{\text{H}}_{\text{3}}}\text{CON}{{\text{H}}_{2}}\] are doubled, then \[L'={{\mu }_{0}}\frac{{{(2N)}^{2}}}{2l}A=2{{\mu }_{0}}\frac{{{N}^{2}}A}{l}=2L\]
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