91影视

(a) The magnetic flux is the amount of magnetic field that flows through a loop of area A. When the magnetic field is perpendicular to the plane's normal, the magnetic flux is given by

${\mathrm{桅}}_m=BA$

However, when the magnetic field forms an angle with the plane, the magnetic flux is given by

${\mathrm{桅}}_{\mathrm{m}}=BA\cos 鈦�蝇t=BA\cos 鈦�\left(2蟺ft\right)$

Where f is the frequency of the rotation. The radius of the semicircle is $r=5\mathrm{cm}$. So, the area of the semicircle is

$蔚=鈭�\frac{d{\mathrm{桅}}_{\mathrm{m}}}{dt}$

$\begin{array}{r}=鈭�\frac{dBA\cos 鈦�\left(2蟺ft\right)}{dt}\\ =鈭�BA\frac{d\cos 鈦�\left(2蟺ft\right)}{dt}\\ =2蟺fBA(\sin 鈦�2蟺ft)\end{array}$

Ohm's law is used to calculate the induced current through the coil, as shown in the following equation.

$I_{\text{induced}}=\frac{蔚}{R}=\frac{2蟺fBA(\sin 鈦�2蟺ft)}{R}$

Now, we plug the values for B, R and A into equation (2) to get$I_{\text{induced in terms of}}t$and f

$I_{\text{induced}}=\frac{2蟺fBA(\sin 鈦�2蟺ft)}{R}$

$\begin{array}{r}=\frac{2蟺\left(0.20\mathrm{T}\right)\left(3.9脳{10}^{鈭�3}\mathrm{m}\right)f\sin 鈦�\left(2蟺ft\right)}{1\mathrm{惟}}\\ =\left(4.9脳{10}^{鈭�3}\right)f\sin 鈦�\left(2蟺ft\right)\end{array}$

(b)We are given the bulb's power and resistance, so we can use these values to calculate the induced current or maximum current for the bulb using the relation. $\left(P=I^{2}R\right)$

$I_{\text{induced}}=I_{max}=\sqrt{\frac{P}{R}}=\sqrt{\frac{4\mathrm{W}}{1\mathrm{惟}}}=2\mathrm{A}$

When the semicircle is perpendicular to the magnetic field, the maximum current is induced. so the term $\sin \left(2蟺ft\right)=\sin 90掳=1$. From our result in part (a), the maximum current will be

$I_{max}=\left(4.9脳{10}^{鈭�3}\right)f$

Solve this equation for f and plug the value of $I_{\max }$

$f=\frac{I_{max}}{4.9脳{10}^{鈭�3}}=\frac{2\mathrm{A}}{4.9脳{10}^{鈭�3}}=408\mathrm{Hz}$