Chapter 30 Problem 30

Problem

Two tightly wound solenoids have the same length and circular cross-sectional area. But solenoid 1 uses wire that is 1.5 times as thick as solenoid 2.

(a) What is the ratio of their inductances?

(b) What is the ratio of their inductive time constants (assuming no other resistance in the circuits)?

Solution

The magnetic field inside a solenoid (ignoring end effects) is constant: $B=m_{0}nI$ where $n=N/l$. The flux is $\Phi _{B}=BA=m_{0}NIA/l$, so

$L={\frac {N\Phi _{B}}{I}}={\frac {m_{0}N^{2}IA}{l}}$

(a)

Both solenoids have the same area and the same length. Because the wire in solenoid 1 is 1.5 times as thick as the wire in solenoid 2, solenoid 2 will have 1.5 times the number of turns as solenoid 1.

${\frac {L_{2}}{L_{1}}}={\frac {\frac {m_{0}N_{2}^{2}IA}{l}}{\frac {m_{0}N_{1}^{2}IA}{l}}}=\left({\frac {N_{2}}{N_{1}}}\right)^{2}=1.5^{2}=2.25$

(b)

To find the ratio of the time constants, both the inductance and resistance ratios need to be known. Since solenoid 2 has 1.5 times the number of turns as solenoid 1, the length of wire used to make solenoid 2 is 1.5 times that used to make solenoid 1, or $l_{\textrm {wire2}}=1.5l_{\textrm {wire1}}$ , and the diameter of the wire in solenoid 1 is 1.5 times that in solenoid 2, or $d_{\textrm {wire1}}=1.5d_{\textrm {wire2}}$. Use this to find their relative resistances, and then the ratio of time constants.

${\frac {R_{1}}{R_{2}}}={\frac {\frac {\rho l_{\textrm {wire1}}}{A_{\textrm {wire1}}}}{\frac {\rho l_{\textrm {wire2}}}{A_{\textrm {wire2}}}}}={\frac {\frac {l_{\textrm {wire1}}}{\pi (0.5d_{\textrm {wire1}})^{2}}}{\frac {l_{\textrm {wire2}}}{\pi (0.5d_{\textrm {wire2}})^{2}}}}={\frac {l_{\textrm {wire1}}}{l_{\textrm {wire2}}}}\left({\frac {d_{\textrm {wire2}}}{l_{\textrm {wire1}}}}\right)^{2}$