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Question: A boat is moving due east in a region where the earth’s magnetic field is \(5\cdot 0\times {{10}^{-5...

A boat is moving due east in a region where the earth’s magnetic field is 5⋅0×10−5 NA−1m−15\cdot 0\times {{10}^{-5}}\text{ N}{{\text{A}}^{-1}}{{\text{m}}^{-1}} due north horizontal. The boat carries a vertical aerial 2 m2\text{ m} long. If the speed of the boat is1⋅50 m−11\cdot 50\text{ }{{\text{m}}^{-1}} , the magnitude of the induced emf in the wire of aerial is:
A.1 mV1\text{ mV}
B.0⋅75 mV0\cdot 75\text{ mV}
C.0⋅50 mV0\cdot 50\text{ mV}
D.0⋅15 mV0\cdot 15\text{ mV}

Explanation

Solution

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⋅0×10−5NA−1m−15\cdot 0\times {{10}^{-5}}\text{N}{{\text{A}}^{-1}}{{\text{m}}^{-1}}
l = 2m2\text{m}
Iv =1⋅50 ms−11\cdot 50\text{ m}{{\text{s}}^{-1}}
Putting all the values
e = Blv
=5 ⋅0×10−5NA−1m−1×2m×1⋅50ms−1 =5⋅0×10−5×3 =15×10−5V 1V=1000mv \begin{aligned} & =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{aligned}
Therefore it will become =0⋅15 mV=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