Question

A 10m wire kept in east-west falling with velocity \[5\,m\,{{s}^{-1}}\] perpendicular to the field \[0.3\times {{10}^{-4}}\,Wb\,{{m}^{-2}}\] . The induced emf across the terminal will be-
(A). \[0.15\,V\]
(B). \[1.5mV\]
(C). \[1.5V\]
(D). \[15\,V\]

Answer 1

Hint: When the wire is falling freely, the magnetic field of the Earth exerts a force on its electrons due to which they move in different directions and a potential difference is developed across its ends. The induced emf is maximum when the field is perpendicular to the length of the wire.

Formula used: \[e=Blv\sin \theta \]

Complete step by step answer:
When the wire is falling perpendicular to the Earth’s magnetic field, the magnetic flux associated with it changes continuously due to which a potential difference develops across its ends, this is called induced emf. The induced emf is given by-
\[e=Blv\sin \theta \] - (1)
Here, \[e\] is induced emf
\[B\] is the magnitude of magnetic field
\[l\] is the length of wire
\[v\] is the velocity of the wire
\[\theta \] is the angle between the magnetic field and length of the wire
 Substituting given values in eq (1), we get,
\[\begin{align}
  & e=0.3\times {{10}^{-4}}\,Wb\,{{m}^{-2}}\times 10m\times 5m{{s}^{-1}} \\
 & \Rightarrow e=15\times {{10}^{-4}}V \\
 & \therefore e=1.5mV \\
\end{align}\]
The value of emf induced between the points of the wire is\[1.5mV\].

So, the correct answer is “Option B”.

Additional Information: When a current is made to pass through a conductor, it develops a magnetic field around it; this is called the magnetic effect of current. This is also called electromagnetism and it has various important applications.

Note: Magnetic flux associated with an area is the number of magnetic lines passing through it. As the number of magnetic lines varies, i.e. the flux changes it induces a potential difference in the conductor due to which current flows. For the induction of emf, the magnetic field must be in the same direction as the area vector.