A boat is moving due East in a region where the earth's magnetic field is 5.0 x 10^-5 N A^-1 m^-1 du

Q: Why did we take the angle as 90 degrees?

The boat is moving East, and the magnetic field is pointing North. East and North are perpendicular directions on the horizontal plane, so the angle between the velocity vector and the magnetic field vector is exactly $90^\circ$.

Q: What if the aerial was laid flat on the boat pointing East?

If the aerial was pointing East, it would be parallel to the velocity vector. The cross product $\vec{v} \times \vec{B}$ points vertically upwards. Since the length vector would be perpendicular to this cross product, the dot product $(\vec{v} \times \vec{B}) \cdot \vec{l}$ would be zero, resulting in zero induced EMF.

Q: Why didn't we use the vertical component of the Earth's magnetic field?

The question explicitly states that the magnetic field in this region is horizontal and points due North. Therefore, there is no vertical component to consider in this specific problem setup.

Q: How do we know which end of the aerial is positive?

You can use Fleming's Right-Hand Rule or the cross product $\vec{v} \times \vec{B}$. With velocity East ($\hat{i}$) and magnetic field North ($\hat{j}$), the force on positive charges $\hat{i} \times \hat{j} = \hat{k}$ is upwards. So, the top of the aerial becomes positively charged.

Q: I got $15 \times 10^{-5}$, how did it become $0.15$ mV?

To convert to millivolts, you need the power of 10 to be $-3$. You can rewrite $15 \times 10^{-5}$ as $15 \times 10^{-2} \times 10^{-3}$. Since $15 \times 10^{-2} = 0.15$, the value becomes $0.15 \times 10^{-3}$ V, which is $0.15$ mV.

Question

A boat is moving due East in a region where the earth's magnetic field is $5.0 	imes 10^{-5} 	ext{ N A}^{-1} 	ext{ m}^{-1}$ due North and horizontal. The boat carries a vertical aerial 2m long. If the speed of the boat is $1.50 	ext{ ms}^{-1}$, the magnitude of the induced emf in the wire of aerial is

Options:

(A) 0.75 mV

(B) 0.50 mV

(C) 0.15 mV

(D) 1 mV

Sigma Percentile · Accepted Answer

Imagine you are standing on a boat, cruising smoothly due East across the water. The Earth's magnetic field in this region is pointing due North. On your boat, there is a vertical aerial (an antenna) standing straight up into the sky.

This is a classic 3D physics problem, and visualizing the setup is half the battle won. We have three distinct directions: the boat's velocity is along the East ($x$-axis), the magnetic field is along the North ($y$-axis), and the aerial is Vertical ($z$-axis).

### The Master Equation: Motional EMF

Whenever a conductor moves through a magnetic field, it cuts through the magnetic lines of force. This action induces an electromotive force (EMF) across the ends of the conductor.

The fundamental formula for this motional EMF is given by the scalar triple product, which simplifies to:
$$e = B l v \sin(	heta)$$
Here, $B$ is the magnetic field strength, $l$ is the length of the conductor, $v$ is its velocity, and $	heta$ is the angle between the velocity vector and the magnetic field vector.

### The Magic of Orthogonality

Let's look closely at our specific setup. The velocity $\vec{v}$ is East, the magnetic field $\vec{B}$ is North, and the length vector $\vec{l}$ is vertical.

Notice how all three vectors are mutually perpendicular to each other! Because the velocity and the magnetic field are at a perfect $90^\circ$ angle, $	heta = 90^\circ$. Since $\sin(90^\circ) = 1$, our formula beautifully simplifies to:
$$e = B l v$$
This is the maximum possible induced EMF for these given values, as the conductor is cutting the magnetic field lines head-on.

### Final Calculation

Now, let's substitute the given values into our simplified equation. We know:
- $B = 5.0 	imes 10^{-5} 	ext{ T}$
- $l = 2 	ext{ m}$
- $v = 1.50 	ext{ m/s}$

Plugging these in, we get:
$$e = (5.0 	imes 10^{-5}) 	imes (2) 	imes (1.50)$$
$$e = 10.0 	imes 10^{-5} 	imes 1.50$$
$$e = 15.0 	imes 10^{-5} 	ext{ V}$$

To match the options provided in the question, we need to convert this result into millivolts ($	ext{mV}$). We can rewrite the expression as:
$$e = 0.15 	imes 10^{-3} 	ext{ V}$$
Since $10^{-3} 	ext{ V}$ is exactly $1 	ext{ mV}$, our final induced EMF is:
**$$e = 0.15 	ext{ mV}$$**

This perfectly matches option (c).

**Food for thought:** What if the boat was moving North instead of East? In that scenario, the velocity $\vec{v}$ and the magnetic field $\vec{B}$ would be parallel. The cross product $\vec{v} 	imes \vec{B}$ would be zero, meaning no magnetic flux would be cut by the aerial, and the induced EMF would be exactly zero! Always visualize the vectors before jumping into calculations.

Select Answer:

Visualized Solution

\text{Visualizing the 3D Setup}

\text{Motional EMF Formula}

\text{Orthogonality Condition}

\text{Substituting the Values}

\text{Calculation}

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Motional EMF

The Master Equation

The Magic of Orthogonality

Final Calculation

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