Question

The magnetic field in a region is given by $\overrightarrow{B}=B_0(1+\frac{x}{a})\hat{k}$. A square loop of edge length $d$ is placed with its edge 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. $\frac{v_0{B}_0{d}^2}{a}$
• B. $\frac{v_0{B}_0{d}^3}{a^2}$
• C. $v_0{B}_{0}d$
• D. $2v_0{B}_{0}d$
  
• E. zero

Answer 1

Answer: (A)

$\frac{v_0{B}_0{d}^2}{a}$

$\varphi ={\int }_x^{x+d}B.dA={\int }_x^{x+d}{B}_o(1+\frac{x}{a})ddx=B_{o}d(d+\frac{(x+d{)}^2-x^2}{2a})$
By faraday's law

$ϵ=-\frac{d\varphi }{dt}=\frac{d\varphi }{dx}\frac{dx}{dt}$

$=B_{o}d\frac{2d}{2a}{v}_o$