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Lenz’s Law | Lenz Law | What is Lenz Law

Lenz’s Law | Lenz Law | What is Lenz Law :- When a changing magnetic field induces an electric current in a circuit, that current itself produces a magnetic field. The direction of this induced current is not random — it always acts in such a way as to oppose the change that caused it.
This principle is stated by Lenz’s Law, discovered by the German scientist Heinrich Friedrich Lenz (1804-1865) in 1834. Lenz’s Law is a direct consequence of the law of conservation of energy and plays a vital role in understanding electromagnetic induction.

Statement of Lenz’s Law

(Lenz’s Law | Lenz Law | What is Lenz Law)

The direction of the induced current (or induced emf) is always such that it opposes the change in magnetic flux which produces it.

In other words, nature resists any change in magnetic conditions. If you try to increase the magnetic flux through a coil, the induced current will create a magnetic field that tries to reduce it, and if you try to decrease the flux, the induced current will try to increase it.

Mathematical Expression

(Lenz’s Law | Lenz Law | What is Lenz Law)

Lenz’s Law is represented by the negative sign in Faraday’s Second Law of electromagnetic induction :

$\displaystyle e=-\frac{d\phi }{dt}$

the negative sign shows that the induced emf opposes the change in magnetic flux.

Physical Explanation

(Lenz’s Law | Lenz Law | What is Lenz Law)

Suppose a magnet is moved towards a coil :

The magnetic flux linked with the coil increases.
According to Lenz’s Law, the coil will produce an induced current whose magnetic field opposes the approaching magnet (i.e., it repels the magnet).

If the magnet is moved away from the coil :

The magnetic flux decreases.
The coil will now produce an induced current whose magnetic field attracts the magnet, opposing its motion away.

Thus, in both cases, the induced current acts to oppose the cause of its generation.

Steps to Determine the Direction of the Induced Current

(Lenz’s Law | Lenz Law | What is Lenz Law)

(1). Identify the direction of the external magnetic field.

Make sure the field lines are passing through the loop.

(2). Observe how the magnetic flux is changing.

Check whether the flux through the loop is increasing, decreasing, or remaining constant.

(3). Find the direction of the induced magnetic field that resists the change in flux.

If the flux increases, the induced magnetic field acts opposite to the applied magnetic field.
If the flux decreases, the induced magnetic field acts in the same direction as the applied field.
If the flux remains constant, no induced magnetic field is produced.

(4). Determine the direction of the induced current.

Apply the right-hand rule to find the direction of the current in the loop that produces the induced magnetic field obtained in Step 3.

Lenz’s Law and Conservation of Energy

(Lenz’s Law | Lenz Law | What is Lenz Law)

We have seen in above diagrams (Physical Explanation) that :-

When a bar magnet is moved toward a conducting coil :

Magnetic flux through the coil increases.
The coil responds by inducing a current that creates a magnetic field opposing the approaching magnet (repelling it).

When the magnet is moved away :

Magnetic flux decreases.
The coil induces current in the opposite direction, producing a field that attracts the magnet (again opposing the change).

Thus, the coil always resists the change in magnetic flux.

Relation to Conservation of Energy

Lenz’s Law is a direct consequence of the Law of Conservation of Energy, which states that energy can neither be created nor destroyed — only transformed from one form to another.

If the induced current aided the change in magnetic flux instead of opposing it, the induced emf would continuously increase the magnetic field, creating energy out of nothing — which is impossible.

By opposing the change, Lenz’s Law ensures that :

Work must be done against the induced current to move the magnet or coil.
The mechanical energy spent in doing this work is converted into electrical energy (in the form of induced current and heat).

Thus, energy is conserved — it merely changes form from mechanical to electrical (and thermal).

🔬 Curio Capsules

(1). Induced emf does not depend on the nature of the coil or its resistance, and it exists even in an open circuit.

(2).If the circuit is open (R = ∞), there will be an induced emf, but no current will flow.

(3). Magnitude of induced emf is directly proportional to the relative speed of coil-magnet system, (e ∝ v).

(4). Induced current depends on the resistance of the coil or circuit.

$\displaystyle i=\frac{e}{R}=-\frac{1}{R}\frac{d\phi }{dt}$

(5). The total induced charge in a coil (or circuit) does not depend on the time during which the change in magnetic flux occurs; that is, it is independent of the rate of change of flux or the relative speed of the coil–magnet system.

$\displaystyle i=\frac{dQ}{dt}$

$\displaystyle \Rightarrow dQ=idt=-\frac{1}{R}\frac{d\phi }{dt}\times dt=-\frac{1}{R}d\phi$

$\displaystyle \Rightarrow Q=-\frac{1}{R}\int\limits_{{{\phi }_{1}}}^{{{\phi }_{2}}}{d\phi }=-\frac{1}{R}\left( {{\phi }_{2}}-{{\phi }_{1}} \right)$

$\displaystyle \therefore Q=\frac{\left( {{\phi }_{1}}-{{\phi }_{2}} \right)}{R}$

(6). Induced charge depends on the change in magnetic flux and the coil/circuit properties (e.g., resistance).

Example 1.

(NCERT Example 6.4)

Below figure shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz’s law.

Solution :

Loop (i) — Rectangular loop

Applied magnetic field: Into the page (⊗).
Flux change: As the loop moves into the magnetic region, the flux increases (more field lines pass through the loop).
Induced magnetic field: For increasing flux, the induced field acts opposite to the applied field → i.e. out of the page (•).
Direction of induced current : To produce a magnetic field out of the page, the current must be anticlockwise (counterclockwise) (use the right-hand rule: thumb out of page → fingers curl anticlockwise). Hence, the current flows a → b → c → d → a.

Result (i): Induced current is anticlockwise.

Loop (ii) — Triangular loop

Applied magnetic field: Into the page (⊗).
Flux change: The triangle is moving out of the region, so the magnetic flux through the triangle is decreasing.
Induced magnetic field: To oppose a decrease in flux, the induced field must support the original field — i.e., it points into the page (⊗).
Direction of induced current: Use the right-hand rule : point your thumb into the page; your fingers curl clockwise. Hence the induced current is clockwise. For triangle vertices, the clockwise route is a → c → b → a.

Result (ii): Induced current is clockwise.

Loop (iii) — Irregular loop (moving out of the field)

Applied magnetic field: Into the page (⊗).
Flux change: The irregular loop is leaving the magnetic region, so the magnetic flux through it is decreasing.
Induced magnetic field: To oppose the decrease the induced field must point into the page (⊗).
Direction of induced current: Use the right-hand rule: thumb into the page → fingers curl clockwise. Hence the induced current flows clockwise around the loop. Tracing the labeled points in clockwise order gives a → d → c → b → a.

Result (iii): Induced current is clockwise.

Example 2.

(NCERT Example 6.5)

(a). A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets ?
(b). A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop
(i) when it is wholly inside the region between the capacitor plates
(ii) when it is partially outside the plates of the capacitor ? The electric field is normal to the plane of the loop.
(c). A rectangular loop and a circular loop are moving out of a uniform magnetic field region (as shown in figure) to a field-free region with a constant velocity v. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops.

(d). Predict the polarity of the capacitor in the situation shown in the figure below.

Solution :

(a). No. Regardless of how strong the magnet is, a current is induced only when the magnetic flux through the loop changes.

(b). No current is induced in either case because a changing electric flux cannot produce an induced current; only a changing magnetic flux can. Faraday’s law applies to changing magnetic flux, not electric flux.

(c). The induced emf is constant in the rectangular loop because its straight edge moves uniformly across the magnetic field boundary, causing the area (and hence flux) within the field to decrease at a constant rate. In contrast, for the circular loop, the enclosed area decreases non-uniformly as it moves out, so the induced emf is not constant.

(d). When the magnet moves toward the loop along its axis, the magnetic flux through the loop increases. According to Lenz’s law, the induced current opposes this increase. Using the right-hand rule, the direction of induced current is such that plate A of the capacitor becomes positive and plate B becomes negative.

Example 3.

A closed coil of copper having 10 turns and dimensions 1m x 1m is free to rotate about an axis. The coil is placed perpendicular to a magnetic field of 0.10 Wb/m². It is rotated through 180º in 0.01 second. The induced emf and induced current in the coil will respectively be – (Resistance of the coil is 2.0 Ω)

(A) 200 V, 100A

(B) 100 V, 200 A

(D) 200 V, 200 A

Solution :

Change is flux linked with the coil on rotating it through 180º will be :

$\displaystyle \Delta \phi =NBA-(-NBA)=2NBA=2\times 10\times 0.10\times \left( 1\times 1 \right)=2Wb$

Induced emf :

$\displaystyle e=-\frac{\Delta \phi }{\Delta t}=-\frac{2}{0.01}=-200Volt$

The coil is closed and has a resistance of 2.0 Ω, therefore induced current :

$\displaystyle \left| i \right|=\frac{e}{R}=\frac{200}{2}=100Amp.$

Hence option (A) is correct.

Example 4.

In above question find the amount of electric charge that flows through the coil.

Solution :

$\displaystyle Q=\frac{\left( {{\phi }_{1}}-{{\phi }_{2}} \right)}{R}=\frac{\left| \Delta \phi \right|}{R}=\frac{2}{2}=1C$

Example 5.

When the resistance in the primary circuit is varied as shown in the figure, what will be the direction of the induced current in the resistor of the secondary circuit ?

Solution :

As the resistance in the primary circuit is varied as shown by arrow, the total resistance of circuit increases → current in the primary decreases → magnetic field in the primary (from right to left along the axis) decreases.

According to Lenz’s law, the secondary coil induces a current that opposes this decrease by producing a magnetic field in the same (right-to-left) direction.

Therefore, the induced current in the resistor flows from A to B.

Example 6.

The ends of a search coil having 20 turns, area of cross-section 1 cm² and resistance 2 Ω are connected to a ballistic galvanometer (B.G.) of resistance 40 Ω. If the plane of search coil is inclined at 30° to the direction of a magnetic field of intensity 1.5 Wb/m², coil is quickly pulled out of the field to a region of zero magnetic field, calculate the charge passed through the galvanometer.

Solution :

Initial amount of flux linked with the coil,

$\displaystyle {{\phi }_{1}}=N(\overset{\to }{\mathop{B}}\,.\overset{\to }{\mathop{A}}\,)=NBAcos\theta =NBAcos\left( 90{}^\circ -30{}^\circ \right)=NBAcos60{}^\circ$

$\displaystyle \Rightarrow {{\phi }_{1}}=20\times 1.5\times (1\times {{10}^{-4}})\times \frac{1}{2}=1.5\times {{10}^{-3}}Wb$

when the coil is pulled out, the flux becomes zero,

$\displaystyle {{\phi }_{2}}=0$

So change in flux,

$\displaystyle \Delta \phi =\left| {{\phi }_{2}}-{{\phi }_{1}} \right|=1.5\times {{10}^{-3}}Wb$

Amount of charge passing through the galvanometer,

$\displaystyle Q=\frac{\left| \Delta \phi \right|}{R}=\frac{1.5\times {{10}^{-3}}}{(40+2)}=0.357\times {{10}^{-4}}C$

🔬 Curio Capsules

(1). A search coil is a small coil of wire used to detect or measure a changing magnetic field. It works on Faraday’s law of electromagnetic induction, which states that an emf is induced in the coil whenever the magnetic flux through it changes. The induced emf is proportional to the rate of change of flux, so by measuring it, the strength and direction of the magnetic field can be determined. Search coils are commonly used in laboratories to study electromagnetic induction and in instruments like metal detectors and magnetic field sensors.

(2). A ballistic galvanometer (B.G.) is a sensitive instrument used to measure the total charge that passes through it during a short interval. Unlike an ordinary galvanometer that measures steady current, it measures the impulse of current produced by a transient emf. Its coil has a high moment of inertia, allowing the entire charge to pass before the coil starts moving, and the first deflection of the coil is proportional to the total charge. It is commonly used in experiments on electromagnetic induction to measure induced charge or flux.

Example 7.

The radius of a circular coil is decreasing steadily at a rate of 10^–2 m/s. A constant and uniform magnetic field of magnitude 1.5 × 10^–3 Wb/m² is applied perpendicular to the plane of the coil. Determine the radius of the coil when the induced electromotive force (emf) in the coil is 2 μV.

Solution :

Induced emf,

$\displaystyle e=-\frac{d\phi }{dt}=-\frac{d}{dt}\left( BA \right)=-B\frac{dA}{dt}=-B\frac{d\left( \pi {{r}^{2}} \right)}{dt}=-2\pi Br\frac{dr}{dt}$

$\displaystyle \Rightarrow r=-\left( \frac{e}{2\pi B\left( \frac{dr}{dt} \right)} \right)$

As radius of coil is decreasing, $\displaystyle \frac{dr}{dt}=-{{10}^{-2}}m/s$ . Putting all the values, we get

$\displaystyle r=-\left( \frac{2\times {{10}^{-6}}}{2\pi \times 1.5\times {{10}^{-3}}\times \left( -{{10}^{-2}} \right)} \right)=0.021m=2.1cm$

Example 8.

A current I = 4.35(1 + 2t) × 10^–2 A increases at a steady rate in a long straight wire. A small circular loop of radius 1mm has its plane parallel to the wire and its centre is placed at a distance of 1m from the wire. The resistance of the loop is 9.2 × 10^–4 Ω. Find the magnitude and the direction of the induced current in the loop.

Solution :

Since the small loop’s radius (r = 1mm = 10^-3m) is much smaller than its distance from the wire (d = 1m), we can approximate the magnetic field as constant across the loop and equal to the value at its center. The magnetic field (B) from a long straight wire at a distance 𝑑 is given by :

$\displaystyle B=\frac{{{\mu }_{o}}I}{2\pi d}=\frac{2\times {{10}^{-7}}\times \,I}{1}=2\times {{10}^{-7}}\times I$

The magnetic flux (Φ) passing through the loop,

$\displaystyle \phi =BA=\left( 2\times {{10}^{-7}}\times I \right)\times \left[ \pi \times {{\left( {{10}^{-3}} \right)}^{2}} \right]=2\pi \times {{10}^{-13}}\times I$

Induced emf,

$\displaystyle e=-\frac{d\phi }{dt}=-2\pi \times {{10}^{-13}}\times \frac{dI}{dt}$

$\displaystyle \Rightarrow e=\left( -2\pi \times {{10}^{-13}} \right)\times \left[ 4.35\times (0+2)\times {{10}^{-2}} \right]$

$\displaystyle \Rightarrow \left| e \right|=5.46\times {{10}^{-14}}Volt$

And the induced current in the loop,

$\displaystyle i=\frac{\left| e \right|}{R}=\frac{5.46\times {{10}^{-14}}}{9.2\times {{10}^{-4}}}=5.9\times {{10}^{-11}}\approx 60pA$

As the current in the wire increases, the magnetic flux linked with the loop also increases. According to Lenz’s law, the induced current in the loop will flow in the opposite direction to oppose this change, i.e., anticlockwise.

Example 9.

Two concentric, coplanar circular loops of wire, each having a resistance of 10^–4 Ωm^–1, have diameters of 0.2 m and 2 m. A time-varying potential difference is applied to the larger loop. Calculate the current induced in the smaller loop.

Solution :

The situation described in the question is shown in figure below :

Current in the larger loop,

$\displaystyle {{I}_{2}}=\frac{V(t)}{{{R}_{2}}}=\frac{\left( 4+2.5t \right)}{\left( {{10}^{-4}} \right)\times \left( 2\pi \times \frac{2}{2} \right)}=\frac{\left( 4+2.5t \right)}{2\pi }\times {{10}^{4}}$

Magnetic field produced by this current at the center of the circular loops,

$\displaystyle B=\frac{{{\mu }_{o}}{{I}_{2}}}{2R}=\frac{4\pi \times {{10}^{-7}}}{2\times 1}\times \left[ \frac{\left( 4+2.5t \right)}{2\pi }\times {{10}^{4}} \right]$

$\displaystyle \Rightarrow B=\left( 4+2.5t \right)\times {{10}^{-3}}$

As the central loop is very small compared to the outer loop (r <<R), we can consider the magnetic field to be constant throughout the small loop. Magnetic flux through the smaller loop,

$\displaystyle \phi =B{{A}_{1}}=\left[ \left( 4+2.5t \right)\times {{10}^{-3}} \right]\times \left[ \pi {{(0.1)}^{2}} \right]=\left( 4+2.5t \right)\pi \times {{10}^{-5}}$

Magnitude of induced emf in smaller loop,

$\displaystyle e=\left| \frac{d\phi }{dt} \right|=\left( 0+2.5\times 1 \right)\pi \times {{10}^{-5}}=2.5\pi \times {{10}^{-5}}Volt$

Finally induced current in smaller loop,

$\displaystyle {{I}_{1}}=\frac{e}{{{R}_{1}}}=\frac{2.5\pi \times {{10}^{-5}}}{\left( {{10}^{-4}} \right)\times \left( 2\pi \times \frac{0.2}{2} \right)}=1.25Amp.$

Example 10.

A circular loop of radius r and resistance R is shown in the figure. A time-varying magnetic field given by B = B₀ e^-t is established perpendicular to the plane of the loop. If the key K is closed, calculate the electrical power developed immediately after closing the key.

Solution :

Induced emf,

$\displaystyle e=-\frac{d\phi }{dt}=-\frac{d\left( BA \right)}{dt}=-A\frac{dB}{dt}=-\pi {{r}^{2}}\frac{d}{dt}\left( {{B}_{o}}{{e}^{-t}} \right)$

$\displaystyle \Rightarrow e=-\pi {{r}^{2}}{{B}_{o}}\left[ {{e}^{-t}}\left( -1 \right) \right]=\pi {{r}^{2}}{{B}_{o}}{{e}^{-t}}$

Just after closing the key (t = 0), Induced emf,

$\displaystyle {{e}_{o}}=\pi {{r}^{2}}{{B}_{o}}{{e}^{0}}=\pi {{r}^{2}}{{B}_{o}}$

The electrical power developed in the resistor R just at the instant the key is closed is,

$\displaystyle P=\frac{e_{0}^{2}}{R}=\frac{B_{0}^{2}\,{{\pi }^{2}}\,{{r}^{4}}}{R}$

Example 11.

A magnet is moved in the direction shown by the arrow between two coils, AB and CD, as illustrated in the figure. Determine the direction of the induced current in each coil.

Lenz's Law | Lenz Law | What is Lenz Law - Curio Physics

Solution :

Since the south pole of the magnet is moving away from coil AB, the end B of this coil behaves like a north pole to oppose the motion of the magnet (and end A will behave like south pole). Similarly, the end C of coil CD behaves like a north pole to repel the approaching magnet (and end D will behave like south pole). Hence

For coil AB: When viewed from end A, the induced current in coil AB flows in the clockwise direction.
For coil CD: When viewed from end D, the induced current in coil CD also flows in the clockwise direction.

Example 12.

(a). When the current flowing from A to B is increasing in magnitude, what will be the direction of the induced current in the loop?
(b). If the current is decreasing in magnitude, what will be the direction of the induced current in the loop?
(c). If, instead of the current, an electron moves in the same direction, what will happen ?

Solution :

(a). When the current in wire AB increases, the magnetic flux linked with the loop (directed out of the page) also increases. Hence, an induced current is produced in the clockwise direction to oppose this increase in flux.

By Lenz’s law, the loop produces a magnetic field into the page, opposing the field of AB. The two magnetic fields are opposite in direction in the region between the wire and the loop, resulting in a repulsive force between the conductor AB and the loop. As a result, the loop exerts a repulsive force on the conductor AB, as shown in the figure.

(b). When the current in wire AB decreases, the magnetic flux linked with the loop (directed out of the page) also decreases. Consequently, an induced current is produced in the anticlockwise direction to oppose this change by increasing the outward flux. As a result, the loop attracts the conductor AB, as shown in the figure.

By Lenz’s law, when the current in wire AB decreases, the loop produces a magnetic field out of the page to oppose the decrease in flux. The magnetic fields of the loop and the wire are therefore in the same direction in the region between them, resulting in an attractive force between the conductor AB and the loop. As a result, the loop attracts the conductor AB, as shown in the figure.

(c). If an electron moves from left to right, the magnetic flux linked with the loop (directed into the page) will first increase and then decrease as the electron passes by. Consequently, the induced current in the loop will be anticlockwise initially and will then reverse direction (becoming clockwise) as the electron moves past the loop.

Example 13.

When a small piece of wire passes between the poles of a horse-shoe magnet in 0.1 s, an emf of 4 × 10^-3 Volt is induced in it. Calculate the magnetic flux between the poles.

(A) 4 × 10^-3 Wb

(B) 4 × 10^-5 Wb

(D) 4 × 10^-6 Wb

Solution :

Magnitude of induced emf,

$\displaystyle e=\left| \frac{\Delta \phi }{\Delta t} \right|$

$\displaystyle \Rightarrow \Delta \phi =e\times \Delta t=\left( 4\times {{10}^{-3}} \right)\times 0.1=4\times {{10}^{-4}}Wb$

Hence option (C) is correct.

Example 14.

In a region of gravity free space, there exists a non-uniform magnetic field $\displaystyle \vec{B}=-{{B}_{0}}{{x}^{3}}\hat{k}$ . A uniform conductor AB of length L and mass m is placed with its end A at origin such that it extends along +x-axis. Find the initial acceleration of the centre of mass and that of end A when a current i flows from A to B.

Solution :

Consider a small section of length at a distance from the left end (end A); the force on this element is :

$\displaystyle d\vec{F}=i\left[ dx\hat{i}\times \left( -{{B}_{0}}{{x}^{3}} \right)\hat{k} \right]={{B}_{0}}i{{x}^{3}}dx\hat{j}$

The force acting on the entire rod is given by :

$\displaystyle \vec{F}=\int{d\vec{F}}=\int_{0}^{L}{{{B}_{0}}i{{x}^{3}}dx\hat{j}=\frac{{{B}_{0}}i{{L}^{4}}}{4}}\hat{j}$

Hence acceleration of the centre-of-mass :

$\displaystyle {{{\vec{a}}}_{com}}=\frac{{\vec{F}}}{m}=\frac{{{B}_{0}}i{{L}^{4}}}{4m}\hat{j}$

Now to determine the acceleration of a point on the rod, we first calculate the angular acceleration of the rod about its center of mass.

The torque acting on the element due to $\displaystyle d\vec{F}$ about the center of mass :

$\displaystyle d\vec{\tau }=\vec{r}\times d\vec{F}=\left( x-\frac{L}{2} \right)\widehat{i}\times \left( {{B}_{0}}i{{x}^{3}}dx\hat{j} \right)$

$\displaystyle \Rightarrow d\vec{\tau }=\left( x-\frac{L}{2} \right){{B}_{0}}i{{x}^{3}}dx\hat{k}$

Net torque on the rod,

$\displaystyle \tau =\int\limits_{0}^{L}{{{B}_{o}}i}\hat{k}\left( {{x}^{4}}-{{x}^{3}}\frac{L}{2} \right)dx$

$\displaystyle \Rightarrow \vec{\tau }={{B}_{0}}i\hat{k}\left( \frac{{{L}^{5}}}{5}-\frac{{{L}^{5}}}{8} \right)=\frac{3{{B}_{0}}i{{L}^{5}}}{40}\hat{k}$

Moment of inertia of a uniform rod about its centre $\displaystyle I=\frac{m{{L}^{2}}}{12}$ . Hence angular acceleration,

$\displaystyle \vec{\alpha }=\frac{{\vec{\tau }}}{I}=\frac{\left( \frac{3{{B}_{0}}i{{L}^{5}}}{40}\hat{k} \right)}{\left( \frac{m{{L}^{2}}}{12} \right)}=\frac{9}{10}\frac{i{{B}_{0}}{{L}^{3}}}{m}\hat{k}$

Now, acceleration of end A is,

$\displaystyle {{{\vec{a}}}_{A}}={{{\vec{a}}}_{com}}+\left( \vec{\alpha }\times {{{\vec{r}}}_{COM\to point\text{ }A}} \right)$

Where $\displaystyle {{{\vec{r}}}_{COM\to point\text{ }A}}$ is position vector of A with respect to centre of mass.

Here $\displaystyle {{{\vec{r}}}_{COM\to point\text{ }A}}=-\frac{L}{2}\widehat{i}$ , hence acceleration of end A :

$\displaystyle {{{\vec{a}}}_{A}}=\frac{{{B}_{0}}i{{L}^{4}}}{4m}\hat{j}+\left( \frac{9}{10}\frac{{{B}_{0}}i{{L}^{3}}}{m}\hat{k} \right)\times \left( -\frac{L}{2}\hat{i} \right)$

$\displaystyle \Rightarrow {{{\vec{a}}}_{A}}=\frac{{{B}_{0}}i{{L}^{4}}}{4m}\hat{j}-\frac{9}{20}\frac{{{B}_{0}}i{{L}^{4}}}{m}\hat{j}$

$\displaystyle \therefore {{{\vec{a}}}_{A}}=\frac{-{{B}_{0}}i{{L}^{4}}}{5m}\hat{j}$

Lenz’s Law | Lenz Law | What is Lenz Law

Statement of Lenz’s Law

(Lenz’s Law | Lenz Law | What is Lenz Law)

Mathematical Expression

(Lenz’s Law | Lenz Law | What is Lenz Law)

Physical Explanation

(Lenz’s Law | Lenz Law | What is Lenz Law)

Steps to Determine the Direction of the Induced Current

(Lenz’s Law | Lenz Law | What is Lenz Law)

Lenz’s Law and Conservation of Energy

(Lenz’s Law | Lenz Law | What is Lenz Law)

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