Answer:
b:1:4
Explanation:
We are given that two solenoid.
Suppose ,the length of each solenoid=l
Cross-sectional area of each solenoid=A
Let , number of turns in in second solenoid,[tex]N_2=N[/tex]
Number of turns in first solenoid,[tex]N_1=2 N[/tex]
We have to find the ratio of self-inductance of the second solenoid to that of the first.
Self- inductance,L=[tex]\frac{\mu_0N^2A}{l}[/tex]
Using the formula
Self- inductance of one solenoid,[tex]L_1=\frac{\mu_0(2N)^2A}{l}=\frac{\mu_04N^2A}{l}[/tex]
Self-inductance of second solenoid,[tex]L_2=\frac{\mu_0N^2A}{l}[/tex]
[tex]\frac{L_2}{L_1}=\frac{N^2}{4N^2}=\frac{1}{4}[/tex]
[tex]L_2:L_1=1:4[/tex]
Hence, option b is true.
Answer:
option (b)
Explanation:
number of turns in the first solenoid, N1 = N
number of turns in the second solenoid, N2 = 2N
Let the current is i and the length is l and the area of crossection of the solenoid is A.
The formula for the self inductance of the solenoid is
[tex]L=\frac{\mu _{0}N^2\times A}{l}[/tex]
It means the self inductance of the solenoid is directly proportional to the square of number of turns.
So, [tex]\frac{L_{1}}{L_{2}}=\frac{N^{2}}{4N^{2}}[/tex]
L1 : L2 = 1 : 4
