A magnetic field B = (B0 y / a)k^ is acting into the paper in the +z direction. B0 and a are positiv

Q: Why don't the vertical arms EH and FG generate EMF?

Motional EMF is generated when a conductor cuts across magnetic field lines. The formula is $e = B l v \sin\theta$, where $\theta$ is the angle between velocity and length. For the vertical arms, their velocity is parallel to their length, so $\theta = 0$, and no EMF is generated.

Q: How do we determine the direction of the induced current?

We use Lenz's Law. As the loop falls, it moves into a region where the magnetic field (pointing into the page) is stronger. This means the inward magnetic flux is increasing. To oppose this increase, the loop induces a current that creates an outward magnetic field. Using the right-hand grip rule, an outward field requires an anti-clockwise current.

Q: Why is the net magnetic force upwards?

The induced current flows anti-clockwise. Using $\mathbf{F} = i(\mathbf{l} \times \mathbf{B})$, the force on the top arm is downwards, and the force on the bottom arm is upwards. Because the magnetic field is stronger at the bottom arm's depth, the upward force is greater than the downward force, resulting in a net upward drag.

Q: What does the constant $K$ represent physically?

The constant $K = \frac{B_0^2 a^2}{mR}$ has units of inverse time ($s^{-1}$). It represents the damping coefficient of the system. A larger $K$ means the magnetic drag is stronger, and the loop will reach its terminal velocity much faster.

Q: Why does the loop reach a terminal velocity instead of accelerating forever?

As the loop falls faster, the motional EMF increases, which increases the induced current, which in turn increases the upward magnetic drag force. Eventually, this upward drag force becomes exactly equal to the downward force of gravity. At this point, the net force is zero, acceleration stops, and the loop falls at a constant terminal velocity.

Question

A magnetic field $B = (B_0 y / a)\hat{\mathbf{k}}$ is acting into the paper in the $+z$ direction. $B_0$ and $a$ are positive constants. A square loop $EFGH$ of side $a$, mass $m$ and resistance $R$ in $x-y$ plane starts falling under the influence of gravity. Note the directions of $x$ and $y$ in the figure. Find
(a) the induced current in the loop and indicate its direction.
(b) the total Lorentz force acting on the loop and indicate its direction.
(c) an expression for the speed of the loop $v(t)$ and its terminal velocity.

Sigma Percentile · Accepted Answer

Imagine a square loop plunging through a magnetic field that gets progressively stronger the deeper it falls. This isn't just a simple free-fall; it's a beautiful interplay of gravity, electromagnetism, and differential equations. Let's break down the physics step by step.

### Analyzing the Setup

We have a square loop $EFGH$ of side $a$ falling in the $x-y$ plane. The magnetic field is directed into the page ($+z$ direction) and is given by $\mathbf{B} = \frac{B_0 y}{a} \hat{\mathbf{k}}$. Notice the $y$ dependence: the field is not uniform. It grows stronger as $y$ increases (i.e., as we go deeper).

As the loop falls with a velocity $v$, its horizontal arms $EF$ and $GH$ slice through the magnetic field lines. This cutting of flux generates a motional EMF. The vertical arms $EH$ and $FG$, however, are moving parallel to their own length, so they do not contribute to the motional EMF.

### The Master Equation for EMF

Let's calculate the EMF generated in the horizontal arms. The top arm $EF$ is at a depth $y$, so the magnetic field there is $B_{EF} = \frac{B_0 y}{a}$. The EMF generated is:

$e_1 = B_{EF} a v = \left(\frac{B_0 y}{a}ight) a v = B_0 y v$

The bottom arm $GH$ is deeper, at a depth $y+a$. The magnetic field there is stronger, $B_{GH} = \frac{B_0 (y+a)}{a}$. The EMF generated is:

$e_2 = B_{GH} a v = \left(\frac{B_0 (y+a)}{a}ight) a v = B_0 (y+a) v$

Because the bottom arm is in a stronger field, it generates a larger EMF. The net EMF driving current around the loop is the difference between the two:

$e_{	ext{net}} = e_2 - e_1 = B_0 a v$

Dividing this net EMF by the loop's resistance $R$ gives us the induced current:

$i = \frac{B_0 a v}{R}$

By Lenz's Law, the induced current must oppose the change in magnetic flux. Since the loop is falling into a region of stronger inward magnetic field, the flux into the page is increasing. To oppose this, the loop will generate its own outward magnetic field, which requires an **anti-clockwise** current.

### The Magnetic Drag Force

Now that we have a current flowing through the loop in a magnetic field, the loop will experience a Lorentz force, $\mathbf{F} = i(\mathbf{l} 	imes \mathbf{B})$.

Applying the right-hand rule, the force on the top arm $EF$ is directed downwards, while the force on the bottom arm $GH$ is directed upwards. Let's calculate their magnitudes:

$F_{EF} = i a B_{EF} = i B_0 y$

$F_{GH} = i a B_{GH} = i B_0 (y+a)$

Since the bottom arm is in a stronger magnetic field, the upward force $F_{GH}$ is greater than the downward force $F_{EF}$. The net magnetic force is therefore directed upwards, acting as a drag against gravity:

$F_m = F_{GH} - F_{EF} = i B_0 a$

Substituting our expression for the current $i$:

$F_m = \left(\frac{B_0 a v}{R}ight) B_0 a = \frac{B_0^2 a^2 v}{R}$

### Final Calculation: The Velocity Profile

We are now ready to set up the equation of motion using Newton's Second Law. The net force on the loop is gravity pulling it down minus the magnetic drag pushing it up:

$m \frac{dv}{dt} = mg - \frac{B_0^2 a^2}{R} v$

To make the math cleaner, let's define a constant $K = \frac{B_0^2 a^2}{mR}$. The equation becomes:

$\frac{dv}{dt} = g - K v$

This is a classic first-order separable differential equation. We can integrate it from an initial velocity of $0$ at $t=0$:

$\int_0^v \frac{dv}{g - Kv} = \int_0^t dt$

Solving this integral yields the velocity as a function of time:

$v(t) = \frac{g}{K} (1 - e^{-Kt})$

As time goes on ($t ightarrow \infty$), the exponential term decays to zero. The loop stops accelerating and reaches a constant **terminal velocity**, $v_T$:

$v_T = \frac{g}{K} = \frac{mgR}{B_0^2 a^2}$

At this terminal velocity, the upward magnetic drag perfectly balances the downward pull of gravity. The loop continues to fall, but its kinetic energy remains constant, and all the lost gravitational potential energy is beautifully dissipated as Joule heating in the resistor.

Visualized Solution

$Visualizing the Setup$

$Motional EMF Concept$

$EMF in Horizontal Arms$

$Net EMF and Induced Current$

$Lorentz Force on Arms$

$Net Magnetic Force$

$Equation of Motion$

$Solving for Velocity$

$Terminal Velocity$

$Energy Considerations$

The Sigma Insight: Motional EMF

Analyzing the Setup

The Master Equation for EMF

The Magnetic Drag Force

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Motional EMF Concept

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EMF in Horizontal Arms

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net EMF and Induced Current

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lorentz Force on Arms

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Magnetic Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Terminal Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Considerations