Question

please help

A conducting loop (of side a = 12 cm) lies on a table at a distance of s = 5.0 cm from a very long, straight wire carrying a current of 1.2 A. If someone pulls the conducting loop away from the wire at 1.5 m/s, what is the value of the induced emf 6.1 times 10^-7 V 1.06 times 10^-7 V 2.1 times 10^-6 V 2.1 times 10^-7 V

Answer 1

side of square of loop is a^2=0.12^2= 0.0144

$B=\frac{\mu _{0}}{2\pi }\frac{I}{S}$

Since $\mu _{0}$ =4pi*10^-7 Tm/A, since I=1.2 A and R=0.1

since flux for moving loop will be

$d\phi =B.dA$   here let dA=a*dS (dS length along moving direction varies and r spacing from wire to loop)

Now intergrating we have    $\phi =\frac{\mu _{0}Ia}{2\pi } \int_{r}^{r+a}\frac{1}{S}.dS=\frac{\mu _{0}Ia}{2\pi }(Ln(r+a)-Ln(r))$

Now EMF we know

$E= - \frac{d\phi }{dt}= - \frac{\mu _{0}Ia}{2\pi }(Ln(r+a)-Ln(r))=\frac{\mu _{0}Ia}{2\pi }\frac{a}{r(r+a)}\frac{dr}{dt}$

here v= dr/dt=velocity =1.5 m/s

thus E=6.0988*10^-7 Volt