A solenoid of radius 1.90 cm has 700 turns and a length of 15.0 cm.(a) Find its inductance.4.66mh(b) Find the rate at which current must change through it to produce an gimf of 90.0 mV. (Enter the magnitude.)
The correct answer and explanation is:
To solve this problem, we need to calculate the inductance of a solenoid and then use it to find the rate of change of current that produces a given induced emf.
Given:
- Radius of solenoid, r=1.90 cm=0.0190 mr = 1.90 \, \text{cm} = 0.0190 \, \text{m}
- Number of turns, N=700N = 700
- Length of solenoid, ℓ=15.0 cm=0.150 m\ell = 15.0 \, \text{cm} = 0.150 \, \text{m}
- Emf, E=90.0 mV=0.090 V\mathcal{E} = 90.0 \, \text{mV} = 0.090 \, \text{V}
(a) Inductance of a solenoid:
The inductance LL of a solenoid is given by the formula: L=μ0N2AℓL = \mu_0 \frac{N^2 A}{\ell}
Where:
- μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7} \, \text{H/m} (permeability of free space)
- A=πr2=π(0.0190)2=1.134×10−3 m2A = \pi r^2 = \pi (0.0190)^2 = 1.134 \times 10^{-3} \, \text{m}^2
Now plug in the values: L=(4π×10−7)⋅7002⋅1.134×10−30.150L = (4\pi \times 10^{-7}) \cdot \frac{700^2 \cdot 1.134 \times 10^{-3}}{0.150} L=(4π×10−7)⋅490000⋅1.134×10−30.150L = (4\pi \times 10^{-7}) \cdot \frac{490000 \cdot 1.134 \times 10^{-3}}{0.150} L≈4π×10−7⋅3703.6≈4.66×10−3 H=4.66 mHL \approx 4\pi \times 10^{-7} \cdot 3703.6 \approx 4.66 \times 10^{-3} \, \text{H} = \boxed{4.66 \, \text{mH}}
(b) Rate of current change to produce emf:
Using the relation: E=L⋅dIdt⇒dIdt=EL\mathcal{E} = L \cdot \frac{dI}{dt} \Rightarrow \frac{dI}{dt} = \frac{\mathcal{E}}{L} dIdt=0.0904.66×10−3≈19.3 A/s\frac{dI}{dt} = \frac{0.090}{4.66 \times 10^{-3}} \approx \boxed{19.3 \, \text{A/s}}
Explanation (300 words):
A solenoid is a tightly wound coil of wire, often used in electromagnetism. When an electric current flows through it, it creates a magnetic field. The property of inductance quantifies a solenoid’s ability to oppose changes in current through the generation of an induced emf (electromotive force). This is a direct consequence of Faraday’s Law of electromagnetic induction, which states that a changing magnetic field induces an emf in a conductor.
In part (a), the inductance of a solenoid depends on its physical characteristics: the number of turns, the cross-sectional area, and the length. The more turns a solenoid has, the stronger the magnetic field it can generate for a given current, hence the greater its inductance. Likewise, a larger area increases the magnetic flux for each turn, while a shorter length leads to a stronger magnetic field, further increasing inductance. We used these principles and the standard formula L=μ0N2AℓL = \mu_0 \frac{N^2 A}{\ell} to find an inductance of 4.66 millihenries (mH).
In part (b), we apply the formula E=LdIdt\mathcal{E} = L \frac{dI}{dt}, which links the induced emf to the rate of change of current through the inductor. This means that to generate a specific emf, we must change the current at a specific rate. Solving for dIdt\frac{dI}{dt} gives us 19.3 A/s. This tells us that the current in the solenoid must increase or decrease at a rate of 19.3 amperes per second to produce an emf of 90.0 mV.
This analysis highlights the foundational role of inductance in electrical circuits, especially in applications involving transformers, motors, and energy storage systems.