Solve.
-A coil of wire rotating in a magnetic field induces a voltage given by e $= 20 \sin \left( \frac { \pi t } { 4 } - \frac { \pi } { 2 } \right)$, where $t$ is time in seconds. Find the smallest positive time to produce a voltage of $10 \sqrt { 2 }$. Round values to the nearest hundredth.
A) $3 \mathrm { sec }$
B) $2.8 \mathrm { sec }$
C) $2.8 \pi \mathrm { sec }$
D) $3 \pi \mathrm { sec }$

Question

Quizplus · Accepted Answer

To find the smallest positive time $t$ that produces a voltage of $10 \sqrt{2}$, we set the given equation equal to $10 \sqrt{2}$:$$10 \sqrt{2} = 20 \sin\left(\frac{\pi t}{4} - \frac{\pi}{2}\right)$$Dividing both sides by 20:$$\frac{10 \sqrt{2}}{20} = \sin\left(\frac{\pi t}{4} - \frac{\pi}{2}\right)$$Simplifying the left side:$$\frac{1}{2} \sqrt{2} = \sin\left(\frac{\pi t}{4} - \frac{\pi}{2}\right)$$Since $\sin(\frac{\pi}{4}) = \frac{1}{2} \sqrt{2}$, we set the inside of the sine function equal to $\frac{\pi}{4}$ (considering the periodic nature of sine and its symmetry around $-\frac{\pi}{2}$):$$\frac{\pi t}{4} - \frac{\pi}{2} = \frac{\pi}{4} \text{ or } \frac{\pi t}{4} - \frac{\pi}{2} = \frac{5\pi}{4}$$Solving for $t$ in both cases:For the first case:$$\frac{\pi t}{4} = \frac{\pi}{4} + \frac{\pi}{2}$$$$\frac{\pi t}{4} = \frac{3\pi}{4}$$$$t = 3$$For the second case:$$\frac{\pi t}{4} = \frac{5\pi}{4} + \frac{\pi}{2}$$This would give a larger value of $t$ than 3, which is not the smallest positive time. Therefore, the smallest positive time $t$ that produces a voltage of $10 \sqrt{2}$ is $3$ seconds.

Solve -A Coil of Wire Rotating in a Magnetic Field =20sin⁡(πt4−π2)= 20 \sin \left( \frac { \pi t } { 4 } - \frac { \pi } { 2 } \right)=20sin(4πt​−2π​)

Solve
-A Coil of Wire Rotating in a Magnetic Field $= 20 \sin \left( \frac { \pi t } { 4 } - \frac { \pi } { 2 } \right)$