Question

The magnetic field in a region is given by $\overrightarrow{B}=B_0\left(\frac{x}{a}\right)\hat{k}$. A sqaure loop of side $d$ is placed with its edges along the $x$ and $y$ axes. The loop is moved with a constant velocity $\overrightarrow{v}=v_0\hat{i}$. The emf induced in the loop is:

• A. $\frac{B_0{v}_o{d}^2}{2a}$
  
• B. $\frac{B_0{v}_o{d}^2}{a}$
  
• C. $\frac{B_0{v}_{o}d}{2a}$
  
• D. $\frac{B_0{v}_0^{2}d}{2a}$
  

Answer 1

The correct option is B $\frac{B_0{v}_o{d}^2}{a}$


Emf in the moving conductor is given as:$$\epsilon =\overrightarrow{v}\times \overrightarrow{B}\cdot \overrightarrow{l}$$ Since $\overrightarrow{v}\times \overrightarrow{B}$ is along the length of the conductor of the loop for two arms, emf induced in these arms willl be, $\epsilon =vBl$.
For other two arms it is perpendicular to the length, so emf in those arms will be zero.

Let assume after some time, loop has moved a distance of $x$.
${\epsilon }_1=v_0\frac{B_0(x+d)}{a}d$
${\epsilon }_2=v_0\frac{B_{0}x}{a}d$
${\epsilon }_{net}={\epsilon }_1-{\epsilon }_2=\frac{B_0{v}_0{d}^2}{a}$