Question

A rectangular wire loop with length a and width b lies in the xy-plane as shown. Within the loop, there is a time dependent magnetic field given by $B = c [( x \cos \omega t ) \hat{ i }+( y \sin \omega t ) \hat{ k }]$ Here, $C$ and $\omega$ are constants. The magnitude of emf induced in the loop as a function of time is

• A. $\left| \frac{ab^2 c}{2} \, \omega \, \cos \, \omega t \right|$
• B. $\left| ab^2 c \, \omega \, \cos \, \omega t \right|$
• C. $\left| \frac{ab^2 c}{2} \, \omega \, \sin \, \omega t \right|$
• D. None of the options

Answer 1

Answer: (A)

Area vector of the given loop,
$A = ab \hat{ k }$
Time dependent magnetic field,
$B = c [( x \cos \omega t ) \hat{ i }+(y \sin \omega t ) \hat{ k }]$
$\therefore $ Magnetic flux through the loop,
$\phi_{n}=A \cdot B$
$=a b \hat{k}[c\{(x \cos \omega t) \hat{i}+(y \sin \omega t) \hat{k}\}]$
$=abc\, y\, \sin \omega t$...(i)
$\because y \leq y \leq b$
$\therefore y_{a v}=\frac{0+b}{2}=\frac{b}{2}$
Putting the value of $y$ in Eq. (i), we get
$\therefore \phi_{ n }= abc \cdot \frac{ b }{2} \cdot \sin \omega t =\frac{ ab ^{2}}{2} c\sin \omega t$
$\therefore $ Induced emf in the loop,
$\varepsilon=\frac{ d }{ dt } \phi_{ n }=\frac{ d }{ dt } \cdot \frac{ ab ^{2} c }{2} \sin \omega t$
Hence, $|\varepsilon|=\left|\frac{ ab ^{2} c }{2} \omega \cos \omega t \right|$