Question

A wire shown in the figure is moving with a velocity of $10\text{m/s}$ in the magnetic field of magnitude $0.3\text{T}$. If the distance between the points $\text{A}$ and $\text{B}$ is $15\text{cm}$. Calculate the emf induced between points $\text{A}$ and $\text{B}$.

• A. $0\text{V}$
• B. $0.9\text{V}$
• C. $1.25\text{V}$
• D. $0.45\text{V}$

Answer 1

The correct option is D $0.45\text{V}$


The emf induced in an irregular shape wire is equal to the emf induced across an imaginary straight wire joining the ends of the given wire.

$\therefore$ Induced emf, $E=Bvl....\left(1\right)$

Given,

$B=0.3\text{T};v=10\text{m/s};l=15\text{cm}=0.15\text{m}$

Substituting the values in equation $\left(1\right)$,

$\Rightarrow E=0.3\times 10\times 0.15=0.45\text{V}$

Hence, $\left(D\right)$ is the correct answer.