13.3: Sheria ya Lenz

Malengo ya kujifunza

Mwishoni mwa sehemu hii, utaweza:

• Tumia sheria ya Lenz kuamua mwelekeo wa emf ikiwa wakati wowote mabadiliko ya magnetic flux
• Tumia sheria ya Faraday na sheria ya Lenz kuamua emf ikiwa katika coil na katika solenoid

Mwelekeo ambao emf inayotokana inaendesha sasa karibu na kitanzi cha waya inaweza kupatikana kupitia ishara hasi. Hata hivyo, kwa kawaida ni rahisi kuamua mwelekeo huu na sheria ya Lenz, iliyoitwa kwa heshima ya mvumbuzi wake, Heinrich Lenz (1804—1865). (Faraday pia aligundua sheria hii, kwa kujitegemea Lenz.) Tunasema sheria ya Lenz kama ifuatavyo:

Sheria ya Lenz

Mwelekeo wa emf inayotokana na sasa karibu na kitanzi cha waya ili kupinga mabadiliko katika flux magnetic ambayo husababisha emf.

Sheria ya Lenz pia inaweza kuchukuliwa katika suala la uhifadhi wa nishati. Ikiwa kusuuza sumaku ndani ya coil husababisha sasa, nishati katika sasa hiyo lazima imetoka mahali fulani. Ikiwa sasa ikiwa husababisha uwanja wa magnetic kupinga ongezeko la uwanja wa sumaku tuliyoingiza, basi hali hiyo ni wazi. Sisi kusukwa sumaku dhidi ya shamba na alifanya kazi kwenye mfumo, na kwamba ilionyesha kama sasa. Kama haikuwa hivyo kwamba uwanja ikiwa inapinga mabadiliko katika flux, sumaku itakuwa vunjwa katika kuzalisha sasa bila kitu baada ya kufanya kazi. Nishati ya uwezo wa umeme ingekuwa imeundwa, kukiuka uhifadhi wa nishati.

Kuamua emf ikiwa\(\epsilon\), wewe kwanza kuhesabu flux magnetic\(\Phi_m\) na kisha kupata\(d\Phi_m/dt\). Ukubwa wa\(\epsilon\) hutolewa na

\[\epsilon = \left|\dfrac{d\Phi_m}{dt}\right|.\]

Hatimaye, unaweza kutumia sheria Lenz kuamua maana ya\(\epsilon\). Hii itaendelezwa kupitia mifano inayoonyesha mkakati wa kutatua matatizo yafuatayo.

Mkakati wa kutatua matatizo: Sheria ya Lenz

Kutumia sheria ya Lenz kuamua maelekezo ya mashamba ya magnetic, mikondo, na emfs:

• Fanya mchoro wa hali ya matumizi katika maelekezo ya kutazama na kurekodi.
• Kuamua mwelekeo wa uwanja wa magnetic uliotumiwa\(\vec{B}\).
• Kuamua kama flux yake ya magnetic inaongezeka au kupungua.
• Sasa tambua mwelekeo wa shamba la magnetic linaloingizwa\(\vec{B}\). Shamba la magnetic linajaribu kuimarisha flux ya magnetic ambayo inapungua au inapinga flux ya magnetic inayoongezeka. Kwa hiyo, shamba la magnetic linaloongeza au linaondoa kwenye uwanja wa magnetic uliotumiwa, kulingana na mabadiliko katika flux ya magnetic.
• Tumia utawala wa mkono wa kulia 2 (RHR-2; tazama Vikosi vya Magnetic na Mashamba) kuamua mwelekeo wa sasa ulioingizwa mimi ambao ni wajibu wa shamba la magnetic linaloingizwa\(\vec{B}\).
• Mwelekeo (au polarity) ya emf iliyosababishwa sasa inaweza kuendesha sasa ya kawaida katika mwelekeo huu.

Hebu tufanye sheria ya Lenz kwenye mfumo wa Kielelezo\(\PageIndex{1a}\). Tunaweka “mbele” ya kitanzi kilichofungwa kama kanda iliyo na sumaku ya bar inakaribia, na “nyuma” ya kitanzi kama kanda nyingine. Kama pole ya kaskazini ya sumaku inakwenda kuelekea kitanzi, flux kupitia kitanzi kutokana na shamba la sumaku huongezeka kwa sababu nguvu ya mistari ya shamba iliyoongozwa kutoka mbele hadi nyuma ya kitanzi inaongezeka. Kwa hiyo sasa inaingizwa katika kitanzi. Kwa sheria ya Lenz, mwelekeo wa sasa unaosababishwa lazima uwe kama shamba lake la magnetic linaelekezwa kwa njia ya kupinga mabadiliko ya mabadiliko yanayosababishwa na uwanja wa sumaku inakaribia. Kwa hiyo, sasa inazunguka ili mstari wake wa magnetic kupitia kitanzi huelekezwa kutoka nyuma hadi mbele ya kitanzi. Kwa RHR-2, weka kidole chako kinachozungumzia dhidi ya mistari ya shamba la magnetic, ambayo inaelekea sumaku ya bar. Vidole vyako vifungia kwa mwelekeo kinyume kama inavyotazamwa kutoka kwenye sumaku ya bar. Vinginevyo, tunaweza kuamua mwelekeo wa sasa ikiwa kwa kutibu kitanzi cha sasa kama umeme unaopinga njia ya kaskazini ya sumaku ya bar. Hii hutokea wakati mtiririko wa sasa unaoonyeshwa kama inavyoonekana, kwa maana uso wa kitanzi karibu na sumaku inakaribia pia ni pole ya kaskazini.

Kielelezo\(\PageIndex{1}\): Mabadiliko katika flux magnetic unasababishwa na sumaku inakaribia induces sasa katika kitanzi. (a) Inakaribia kaskazini pole induces sasa kinyume na heshima na sumaku bar. (b) Pole ya kusini inakaribia inasababisha sasa ya saa kwa heshima na sumaku ya bar.

Sehemu (b) ya takwimu inaonyesha pole ya kusini ya sumaku inayohamia kuelekea kitanzi cha kufanya. Katika kesi hii, kuongezeka kwa njia ya kitanzi kutokana na shamba la sumaku huongezeka kwa sababu idadi ya mistari ya shamba iliyoongozwa kutoka nyuma hadi mbele ya kitanzi inaongezeka. Ili kupinga mabadiliko haya, sasa inaingizwa katika kitanzi ambacho mistari ya shamba kupitia kitanzi huelekezwa kutoka mbele hadi nyuma. Kwa usawa, tunaweza kusema kwamba sasa inapita katika mwelekeo ili uso wa kitanzi karibu na sumaku inakaribia ni pole ya kusini, ambayo kisha inakaribia pole ya kusini ya sumaku. Kwa RHR-2, alama yako ya kidole mbali na sumaku ya bar. Vidole vyako vifunga kwa mtindo wa saa moja, ambayo ni mwelekeo wa sasa ulioingizwa.

Mfano mwingine unaoonyesha matumizi ya sheria ya Lenz unaonyeshwa kwenye Kielelezo\(\PageIndex{2}\). Wakati kubadili kufunguliwa, kupungua kwa sasa kwa njia ya solenoid husababisha kupungua kwa magnetic flux kupitia coils yake, ambayo inasababisha emf katika solenoid. EMF hii inapaswa kupinga mabadiliko (kusitishwa kwa sasa) na kusababisha. Kwa hiyo, emf iliyosababishwa ina polarity iliyoonyeshwa na inaendesha katika mwelekeo wa sasa wa awali. Hii inaweza kuzalisha arc katika vituo vya kubadili kama inafunguliwa.

Kielelezo\(\PageIndex{2}\): (a) solenoid kushikamana na chanzo cha EMF. (b) Kufungua kubadili S hupunguza sasa, ambayo kwa hiyo inasababisha emf katika solenoid. (c) Tofauti tofauti kati ya mwisho wa viboko vilivyoelekezwa huzalishwa kwa kuchochea emf katika coil. Tofauti hii ya uwezo ni kubwa ya kutosha kuzalisha arc kati ya pointi kali.

Zoezi\(\PageIndex{1A}\)

Pata mwelekeo wa sasa ulioingizwa katika kitanzi cha waya kilichoonyeshwa hapa chini kama sumaku inaingia, inapita, na kuacha kitanzi.

Suluhisho

Kwa mwangalizi umeonyesha, sasa inapita mwendo wa saa kama sumaku inakaribia, itapungua hadi sifuri wakati sumaku inazingatia katika ndege ya coil, na kisha inapita kinyume chake kama sumaku inavyoacha coil.

Zoezi\(\PageIndex{1B}\)

Verify the directions of the induced currents in Figure 13.2.2.

Example \(\PageIndex{1A}\): A Circular Coil in a Changing Magnetic Field

A magnetic field \(\vec{B}\) is directed outward perpendicular to the plane of a circular coil of radius \(r = 0.50 \, m\) (Figure \(\PageIndex{3}\)). The field is cylindrically symmetrical with respect to the center of the coil, and its magnitude decays exponentially according to \(B = (1.5T)e^{(5.0s^{-1})t}\), where B is in teslas and t is in seconds. (a) Calculate the emf induced in the coil at the times \(t_1 = 0\), \(t_2 = 5.0 \times 10^{-2}s\), and \(t_3 = 1.0 \, s\). (b) Determine the current in the coil at these three times if its resistance is \(10 \, \Omega\).

Figure \(\PageIndex{3}\): A circular coil in a decreasing magnetic field.

Strategy

Since the magnetic field is perpendicular to the plane of the coil and constant over each spot in the coil, the dot product of the magnetic field \(\vec{B}\) and normal to the area unit vector \(\hat{n}\) turns into a multiplication. The magnetic field can be pulled out of the integration, leaving the flux as the product of the magnetic field times area. We need to take the time derivative of the exponential function to calculate the emf using Faraday’s law. Then we use Ohm’s law to calculate the current.

Solution

1. Since \(\vec{B}\) is perpendicular to the plane of the coil, the magnetic flux is given by \[\Phi_m = B\pi r^2 = (1.5 e^{-5.0 t}T)\pi (0.50 \, m)^2\]\[= 1.2 e^{-(5.0 s^{-1})t} Wb.\] From Faraday’s law, the magnitude of the induced emf is \[\epsilon = \left|\frac{d\Phi_m}{dt}\right| = \left|\frac{d}{dt} (1.2 e^{-(5.0s^{-1})t} Wb)\right| = 6.0 e^{-(5.0s^{-1})t}V.\] Since \(\vec{B}\) is directed out of the page and is decreasing, the induced current must flow counterclockwise when viewed from above so that the magnetic field it produces through the coil also points out of the page. For all three times, the sense of ε is counterclockwise; its magnitudes are \[\epsilon (t_1) = 6.0 V; \, \epsilon (t_2) = 4.7 \, V; \, \epsilon (t_3) = 0040 \, V.\]
2. From Ohm’s law, the respective currents are \[I(t_1) = \frac{\epsilon (t_1)}{R} = \frac{6.0 \, V}{10 \, \Omega} = 0.60 \, A;\]\[I(t_2) = \frac{4.7 \, V}{10 \, \Omega} = 0.47 \, A;\] and \[I(t_3) = \frac{0.040 \, V}{10 \, \Omega} = 4.0 \times 10^{-3} \, A.\]

Significance

An emf voltage is created by a changing magnetic flux over time. If we know how the magnetic field varies with time over a constant area, we can take its time derivative to calculate the induced emf.

Example \(\PageIndex{1B}\): Changing Magnetic Field Inside a Solenoid

The current through the windings of a solenoid with \(n = 2000\) turns per meter is changing at a rate \(dI/dt = 3.0 \, A/s\). (See Sources of Magnetic Fields for a discussion of solenoids.) The solenoid is 50-cm long and has a cross-sectional diameter of 3.0 cm. A small coil consisting of \(N = 20\) closely wound turns wrapped in a circle of diameter 1.0 cm is placed in the middle of the solenoid such that the plane of the coil is perpendicular to the central axis of the solenoid. Assuming that the infinite-solenoid approximation is valid at the location of the small coil, determine the magnitude of the emf induced in the coil.

Strategy

The magnetic field in the middle of the solenoid is a uniform value of \(\mu_0 nI\). This field is producing a maximum magnetic flux through the coil as it is directed along the length of the solenoid. Therefore, the magnetic flux through the coil is the product of the solenoid’s magnetic field times the area of the coil. Faraday’s law involves a time derivative of the magnetic flux. The only quantity varying in time is the current, the rest can be pulled out of the time derivative. Lastly, we include the number of turns in the coil to determine the induced emf in the coil.

Solution

Since the field of the solenoid is given by \(B = \mu_0 nI\), the flux through each turn of the small coil is \[\Phi_m = \mu_0 nI\left(\frac{\pi d^2}{4}\right),\]

where d is the diameter of the coil. Now from Faraday’s law, the magnitude of the emf induced in the coil is

\[\epsilon = \left|N\frac{d\Phi_m}{dt}\right| = \left|N\mu_0 n\frac{\pi d^2}{4} \frac{dI}{dt}\right|\]

\[= 20 (4\pi \times 10^{-7} T \cdot m/s)(2000 \, m^{-1}) \frac{\pi(0.010 \, m)^2}{4} (3.0 \, A/s)\]\[= 1.2 \times 10^{-5} \, V.\]

Significance

When the current is turned on in a vertical solenoid, as shown in Figure \(\PageIndex{4}\), the ring has an induced emf from the solenoid’s changing magnetic flux that opposes the change. The result is that the ring is fired vertically into the air.

Figure \(\PageIndex{4}\): The jumping ring. When a current is turned on in the vertical solenoid, a current is induced in the metal ring. The stray field produced by the solenoid causes the ring to jump off the solenoid.

Note

A demonstration of the jumping ring from MIT.