A solid metal cube of edge length 2 cm is moving in a positive Y-direction at a constant speed of 6

Q: Why do we only consider the length along the X-axis?

Because the magnetic force on the charges is given by $q(\vec{v} \times \vec{B})$. Since $\vec{v}$ is along Y and $\vec{B}$ is along Z, their cross product is along X. Therefore, charges are pushed along the X-axis, creating a potential difference only between the faces perpendicular to the X-axis.

Q: What happens to the other faces of the cube?

There is no magnetic force pushing charges along the Y or Z axes. Therefore, there is no charge accumulation and no potential difference between the faces perpendicular to the Y-axis or the Z-axis.

Q: Why did we convert cm to m?

To get the final answer in Volts (which is an SI unit), all input values must be in SI units. The magnetic field is in Tesla and velocity is in m/s, so the length must be in meters.

Q: Does the shape of the object matter for motional EMF?

The overall shape doesn't matter as much as the 'effective length' perpendicular to both velocity and magnetic field. A wire, a rectangular plate, or a cube will all have the same EMF if their effective length $l$ is the same.

Q: What if the velocity was at an angle?

If the velocity was at an angle, you would need to use the general vector formula $E = |(\vec{v} \times \vec{B}) \cdot \vec{l}|$. You would break the velocity into components and only the component perpendicular to both $\vec{B}$ and $\vec{l}$ would contribute to the EMF.

Question

A solid metal cube of edge length $2	ext{ cm}$ is moving in a positive Y-direction at a constant speed of $6	ext{ m/s}$. There is a uniform magnetic field of $0.1	ext{ T}$ in the positive Z-direction. The potential difference between the two faces of the cube perpendicular to the X-axis is

Options:

(A) $2	ext{ mV}$

(B) $12	ext{ mV}$

(C) $6	ext{ mV}$

(D) $1	ext{ mV}$

Sigma Percentile · Accepted Answer

Imagine you are standing in a vast, invisible ocean of magnetic fields. A solid metal cube is sailing through this ocean, cutting across the magnetic field lines. This is the classic scenario of **Motional EMF**, a beautiful intersection of kinematics and electromagnetism.

### Analyzing the Setup

Let's break down the physical reality of the problem. We have a solid metal cube with an edge length of $a = 2	ext{ cm}$. It is moving with a constant velocity $\vec{v} = 6\hat{j}	ext{ m/s}$ (along the positive Y-axis). The region is permeated by a uniform magnetic field $\vec{B} = 0.1\hat{k}	ext{ T}$ (along the positive Z-axis).

The question asks for the potential difference between the two faces of the cube that are perpendicular to the X-axis. Why specifically these faces? To understand this, we need to look at the microscopic level.

### The Master Equation: Lorentz Force

Inside the metal cube, there is a sea of free electrons. As the cube moves, these electrons move with it. According to the Lorentz force law, a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a magnetic force:

$$\vec{F}_m = q(\vec{v} 	imes \vec{B})$$

Let's find the direction of this force. The velocity is in the $\hat{j}$ direction, and the magnetic field is in the $\hat{k}$ direction. The cross product $\hat{j} 	imes \hat{k}$ gives $\hat{i}$. This means the magnetic force pushes positive charges towards the positive X-face and negative charges (electrons) towards the negative X-face.

This separation of charges creates an electric field $\vec{E}$ inside the cube, pointing from the positive X-face to the negative X-face. This electric field exerts an electric force $\vec{F}_e = q\vec{E}$ that opposes the magnetic force. Equilibrium is reached when these forces balance out:

$$qE = qvB \implies E = vB$$

### Calculating the Motional EMF

The potential difference, or Motional EMF, across a length $l$ is given by the general formula:

$$V = |(\vec{v} 	imes \vec{B}) \cdot \vec{l}|$$

Since $\vec{v}$, $\vec{B}$, and the length vector $\vec{l}$ (which is along the X-axis) are all mutually perpendicular, the formula simplifies beautifully to:

$$V = Bvl$$

Here, $l$ is the distance between the two faces perpendicular to the X-axis. This distance is simply the edge length of the cube, $l = 2	ext{ cm} = 0.02	ext{ m}$.

### Final Calculation

Now, it's just a matter of plugging in the numbers. Don't forget to convert centimeters to meters to maintain SI units!

$$V = (0.1	ext{ T}) 	imes (6	ext{ m/s}) 	imes (0.02	ext{ m})$$

$$V = 0.012	ext{ V}$$

Since the options are given in millivolts, we multiply by $1000$:

$$V = 12	ext{ mV}$$

And there we have it! The potential difference across the X-faces is **$12	ext{ mV}$**.

Always remember, the key to motional EMF problems in 3D is identifying the three mutually perpendicular vectors: velocity, magnetic field, and the effective length across which the charges accumulate.

Select Answer:

Visualized Solution

$Visualizing the 3D Setup$

$Motional EMF Formula$

$Identifying the Effective Length$

$Substituting the Values$

$Calculating the EMF$

$Final Answer$

$The Way Forward$

The Sigma Insight: Motional EMF

Analyzing the Setup

The Master Equation

Calculating the Motional EMF

Final Calculation

Similar Questions

The magnetic field in a region is given by $B = B_{0} (\frac{x}{a}) \hat{k}$ . A square loop of side $d$ is placed with its edges along the X and Y-axes. The loop is moved with a constant velocity $v = v_{0} \hat{i}$ . The emf induced in the loop is

A metal conductor of length $1 m$ rotates vertically about one of its ends at angular velocity $5 rad/s$ . If the horizontal component of earth's magnetic field is $0.2 \times 1 0^{- 4} T$ , then the emf developed between the two ends of the conductor is

A $10 m$ long horizontal wire extends from North-East to South-West. It is falling with a speed of $5.0 ms^{- 1}$ at right angles to the horizontal component of the earth's magnetic field of $0.3 \times 1 0^{- 4} Wb/m^{2}$ . The value of the induced emf in wire is

A metallic square loop $A B C D$ is moving in its own plane with velocity $v$ in a uniform magnetic field perpendicular to its plane as shown in the figure. Electric field is induced

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the 3D Setup

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

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

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

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

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

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