This question involves electromagnetic induction due to motion in a magnetic field. When a conductor moves through a magnetic field, an electromotive force (emf) is induced in it. This is a consequence of Faraday's law, but specifically, for a straight conductor moving perpendicular to a magnetic field, we use the formula for motional emf.

The formula for the magnitude of the motional emf induced in a straight conductor of length $l$ moving with a velocity $v$ perpendicular to a uniform magnetic field $B$ is given by:

$\epsilon =Blv$

However, this formula is only valid when the velocity of the conductor, the magnetic field, and the length of the conductor (which defines the direction of the induced emf) are all mutually perpendicular to each other.

Step 1: Analyze the Given Vectors

• Direction of Boat's Motion (Velocity, $\overrightarrow{v}$): Due East.
• Direction of Earth's Magnetic Field ($\overrightarrow{B}$): Due North and Horizontal.
• Orientation of the Aerial (Length, $\overrightarrow{l}$): Vertical (i.e., perpendicular to the horizontal plane).

Step 2: Determine the Effective Components

The induced emf depends on the component of the velocity that is perpendicular to both the magnetic field and the length of the conductor. Since the aerial is vertical, the effective length vector $\overrightarrow{l}$ is perpendicular to the horizontal plane. The velocity $\overrightarrow{v}$ (East) and the magnetic field $\overrightarrow{B}$ (North) both lie in the horizontal plane. Therefore, the velocity is perpendicular to the magnetic field (East is perpendicular to North).

The motional emf is caused by the magnetic force on the free electrons in the conductor. The force on a charge is given by $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$. For maximum emf, this force must be along the length of the wire. Here, the cross product $\overrightarrow{v}\times \overrightarrow{B}$ (East × North) gives a vector in the downward direction (using the right-hand rule). This direction is parallel to the vertical aerial. Therefore, all conditions for the formula $\epsilon =Blv$ are satisfied.

Step 3: Substitute the Values

Given:
$B=5.0\times {10}^{–5}N{A}^{–1}{m}^{–1}$ (This unit is equivalent to Tesla, T)
$l=2m$
$v=1.50m{s}^{–1}$

Now, calculate the emf:
$\epsilon =Blv=(5.0\times {10}^{–5})\times \left(2\right)\times \left(1.50\right)$

First, multiply the constants: $5.0\times 2\times 1.50=15.0$
Now, combine with the power of ten: $15.0\times {10}^{–5}=1.5\times {10}^{–4}V$

Convert Volts to millivolts (1 V = 1000 mV):
$\epsilon =1.5\times {10}^{–4}\times {10}^{3}mV=1.5\times {10}^{–1}mV=0.15mV$

Final Answer: The magnitude of the induced emf is 0.15 mV.

Related Topics & Formulae

Motional Electromotive Force: This is the emf induced in a conductor moving through a magnetic field. The general vector form of the formula is $\epsilon =\int (\overrightarrow{v}\times \overrightarrow{B})·d\overrightarrow{l}$. For a straight rod in uniform fields, it simplifies to $\epsilon =Blv\mathrm{sin\theta }$, where $\theta$ is the angle between the velocity vector and the magnetic field vector. The maximum emf is induced when the motion is perpendicular to the field.

Faraday's Law of Induction: The more fundamental law states that the induced emf in a closed loop is equal to the negative rate of change of magnetic flux through the loop: $\epsilon =-\frac{d{\Phi }_B}{dt}$. The motional emf scenario is a specific application of this law.

Magnetic Force on a Moving Charge: The underlying principle for motional emf is the magnetic force acting on the free electrons inside the moving conductor, given by $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$. This force causes the electrons to accumulate at one end of the conductor, creating an electric field and thus an emf.