An inductor is formed by winding N turns of a thin conducting wire into a circular loop of radius a . The inductor loop is in the x-y plane with its center at the origin and connected to a resistor R, as shown in Fig. 6-3. In the presence of a magnetic field B=B_(0)(hat (y)2+hat (z)3)sinomega t where a is the angular frequency, find (a) the magnetic flux linking a single turn of the inductor, (b) the transformer emf given that N=10, B_(0)=0.2T, a=10cm and omega =10^3rad/s (c) the polarity of V_(cmf)^tr at t=0 and (d) the induced current in the circuit for R=1kOmega (assume the wire resistance to be much smaller than R)

(a) The magnetic flux linking a single turn of the inductor is given by:<br /><br />$\Phi = B_0 A \cos(\omega t)$<br /><br />where $B_0$ is the magnetic field strength, $A$ is the area of the loop, and $\omega$ is the angular frequency.<br /><br />(b) The transformer emf is given by:<br /><br />$V_{cmf}^{tr} = -N \frac{d\Phi}{dt}$<br /><br />Substituting the expression for $\Phi$ from part (a), we get:<br /><br />$V_{cmf}^{tr} = -N B_0 A \omega \sin(\omega t)$<br /><br />(c) The polarity of $V_{cmf}^{tr}$ at $t$ can be determined by substituting $t=0$ into the expression for $V_{cmf}^{tr}$ from part (b). Since $\sin(\omega t)$ is zero at $t=0$, the emf is also zero at that time. Therefore, the polarity of $V_{cmf}^{tr}$ at $t=0$ is indeterminate.<br /><br />(d) The induced current in the circuit can be found using Ohm's law:<br /><br />$I = \frac{V_{cmf}^{tr}}{R}$<br /><br />Substituting the expression for $V_{cmf}^{tr}$ from part (b), we get:<br /><br />$I = \frac{-N B_0 A \omega \sin(\omega t)}{R}$<br /><br />Substituting the given values, we get:<br /><br />$I = \frac{-10 \times 0.2 \times 0.1^2 \times 10^3 \sin(10^3 t)}{1000}$<br /><br />Simplifying further, we get:<br /><br />$I = -2 \times 10^{-3} \sin(10^3 t)$<br /><br />Therefore, the induced current in the circuit is $-2 \times 10^{-3} \sin(10^3 t)$ A.