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

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,

& 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{aligned}$$ The value of emf induced between the points of the wire is$$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.