Question

3. The magnetic field shown in Figure 4 directed into the paper. The initial diameter quickly pulled taut, and the kink shrinks to f the kinkc is magnitude igure 4 the average voltage induced between endadiameter ofis 2.00cmde of 25.0 mT nk is undisturbed, but the magnetic field increases to a diameter of zero in 50.0 ms. Determine pulled a induced between endpoints 4 and B. Include the polarity. (b) mT in 4.00 the ki average voltage across terminals A and B, including to 100 polarity, during this period. (15) 2.00 cm Figure 4

Answer 1

Magnetic flux $\phi=AB$ ($A$ is area of kink , $B$ is magnetic field)

Induced voltage $\mathcal{E}=-\frac{d\phi}{dt}$

a) Diameter of kink is decreased to zero from $2\,cm$

$A_1= \frac{\pi d^2}{4}=\frac{\pi (0.02)^2}{4}=0.000314\,m^2$

$A_2=0\,m^2$

$B=25\,mT=25*10^{-3}\,T$

time taken $t=50\,ms=50*10^{-3}\,s$

${\mathcal{E}=-\frac{d\phi}{dt}=-B\frac{dA}{dt}=-25*10^{-3}*\frac{0-0.000314}{50*10^{-3}}=0.157*10^{-3}\,V=0.157\,mV}$

Point A is at a higher potential compared to point B.

b)

Magnetic field increases from $B_1=25\,mT=25*10^{-3}\,T$ to $B_2=100\,mT=100*10^{-3}\,T$

Area if kink $A= \frac{\pi d^2}{4}=\frac{\pi (0.02)^2}{4}=0.000314\,m^2$

time taken $t=50\,ms=50*10^{-3}\,s$

${\mathcal{E}=-\frac{d\phi}{dt}=-A\frac{dB}{dt}=-0.000314*\frac{(100-25)*10^{-3}}{50*10^{-3}}=-0.471*10^{-3}\,V=-0.471\,mV}$

Point B is at a higher potential compared to point A.