Question

A boat is moving due east in a region where the earth’s magnetic field is $5\cdot 0\times {{10}^{-5}}\text{ N}{{\text{A}}^{-1}}{{\text{m}}^{-1}}$ due north horizontal. The boat carries a vertical aerial $2\text{ m}$ long. If the speed of the boat is$1\cdot 50\text{ }{{\text{m}}^{-1}}$ , the magnitude of the induced emf in the wire of aerial is:
A.$1\text{ mV}$
B.$0\cdot 75\text{ mV}$
C.$0\cdot 50\text{ mV}$
D.$0\cdot 15\text{ mV}$

Answer 1

Hint: An emf induced by motion relative to a magnetic field is called a motional emf. Thus is represented by equation
e = LVB
L = length of object
V = speed of object
B = magnetic field
S.I unit = Volts

Complete answer:
As we know, induced emf is
e = Blv
As per in our question
B = $5\cdot 0\times {{10}^{-5}}\text{N}{{\text{A}}^{-1}}{{\text{m}}^{-1}}$
l = $2\text{m}$
Iv =\[1\cdot 50\text{ m}{{\text{s}}^{-1}}\]
Putting all the values
e = Blv
$\begin{align}
  & =5~\cdot 0\times {{10}^{-5}}\text{N}{{\text{A}}^{-1}}{{\text{m}}^{-1}}\times 2\text{m}\times \text{1}\cdot 50\text{m}{{\text{s}}^{-1}} \\
 & =5\cdot 0\times {{10}^{-5}}\times 3 \\
 & =15\times {{10}^{-5}}\text{V} \\
 & 1\text{V}=1000\text{mv} \\
\end{align}$
Therefore it will become $=0\cdot 15\text{ mV}$

So, option (D) is correct.

Note:
Basic cause of induced emf is the change of magnetic flux linked with a closed circuit. We can increase the induced emf by increasing the no. of turns of wire in the coil. While solving numerical all units should be in S.I units