Question

A $10m$ wire kept in the east-west direction is falling with a velocity of $5m/sec$ perpendicular to the field $0.3\times {10}^{-4}Wb/m^{{}_2}$. The induced $emf$ across the terminal will be

• A. $0.15V$
• B. $1.5mV$
• C. $1.5V$
• D. $15.0V$

Answer 1

The correct option is B

$1.5mV$

Step 1. Given data

Length of wire,$l=10m$

The velocity of the falling wire, $v=5m/s$

Magnetic field is $B=0.3\times {10}^{-4}Wb/m^{{}_2}$

Step 2. Finding induced $emf$

The induced $emf$in the wire due to magnetic field is given by

$e=Blv\sin \left(\theta \right)\mathrm{where}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{motion}\mathrm{and}\mathrm{the}\mathrm{magnetic}\mathrm{field}$

As the direction of motion is perpendicular to the direction of the magnetic field, thus,

$e\mathit{=}Blv\sin \left({90}^{\circ }\right)$

$$\begin{gathered} e=Blv \\ \Rightarrow e=0.3\times {10}^{-4}\times 10\times 5 \\ \Rightarrow e=1.5\times {10}^{-3}V \\ =1.5mV \end{gathered}$$

Hence, option B is correct