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Complete Question
A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .
Part A. Calculate the coil's self-inductance.
Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.
Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?
Answer:
Part A
[tex]L = 0.000863 \ H[/tex]
Part B
[tex]\epsilon = 0.863 \ V[/tex]
Part C
From terminal a to terminal b
Explanation:
From the question we are told that
The number of turns is [tex]N = 590 \ turns[/tex]
The cross-sectional area is [tex]A = 6.20 cm^2 = 6.20 *10^{-4} \ m[/tex]
The radius is [tex]r = 5.0 \ cm = 0.05 \ m[/tex]
Generally the coils self -inductance is mathematically represented as
[tex]L = \frac{ \mu_o N^2 A }{2 \pi * r }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values
[tex]L = \frac{ 4\pi * 10^{-7} * 590^2 6.20 *10^{-4} }{2 \pi * 0.05 }[/tex]
[tex]L = \frac{ 2 * 10^{-7} * 590^2 6.20 *10^{-4} }{ 0.05 }[/tex]
[tex]L = 0.000863 \ H[/tex]
Considering the Part B
Initial current is [tex]I_1 = 5.00 \ A[/tex]
Current at time t is [tex]I_t = 3.0 \ A[/tex]
The time taken is [tex]\Delta t = 3.00 ms = 0.003 \ s[/tex]
The self-induced emf is mathematically evaluated as
[tex]\epsilon = L * \frac{\Delta I}{ \Delta t }[/tex]
=> [tex]\epsilon = L * \frac{ I_1 - I_t }{ \Delta t }[/tex]
substituting values
[tex]\epsilon = 0.000863 * \frac{ 5- 2 }{ 0.003 }[/tex]
[tex]\epsilon = 0.863 \ V[/tex]
The direction of the induced emf is from a to b because according to Lenz's law the induced emf moves in the same direction as the current
This question involves the concepts of the self-inductance, induced emf, and Lenz's Law
A. The coil's self-inductance is "0.863 mH".
B. The self-induced emf in the coil is "0.58 volts".
C. The direction of the induced emf is "from b to a".
A.
The self-inductance of the coil is given by the following formula:
[tex]L=\frac{\mu_oN^2A}{2\pi r}[/tex]
where,
L = self-inductance = ?
[tex]\mu_o[/tex] = permeability of free space = 4π x 10⁻⁷ N/A²
N = No. of turns = 590
A = Cross-sectional area = 6.2 cm² = 6.2 x 10⁻⁴ m²
r = radius = 5 cm = 0.05 m
Therefore,
[tex]L=\frac{(4\pi\ x\ 10^{-7}\ N/A^2)(590)^2(6.2\ x\ 10^{-4}\ m^2)}{2\pi(0.05\ m)}[/tex]
L = 0.863 x 10⁻³ H = 0.863 mH
B.
The self-induced emf is given by the following formula:
[tex]E=L\frac{\Delta I}{\Delta t}\\\\[/tex]
where,
E = self-induced emf = ?
ΔI = change in current = 2 A
Δt = change in time = 3 ms = 0.003 s
Therefore,
[tex]E=(0.000863\ H)\frac{2\ A}{0.003\ s}[/tex]
E = 0.58 volts
C.
According to Lenz's Law, the direction of the induced emf always opposes the change in flux that causes it. Hence, the direction of the induced emf will be from b to a.
Learn more about Lenz's Law here:
https://brainly.com/question/12876458?referrer=searchResults