A flat coil is wrapped with 200 turns of very thin wire on a square frame with sides 18 cm long.A uniform magnetic field is applied perpendicular to the plane of the coil.If the magnitude of the $\vec { B }$ field changes uniformly from 0.50 T to 0.00 T in 8.0 s,find the emf induced in the coil.

Question

Quizplus · Accepted Answer

The emf induced in the coil can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf ($\epsilon$) in a coil is equal to the negative rate of change of magnetic flux ($\Phi_B$) through the coil. The formula is given by $\epsilon = -N \frac{\Delta \Phi_B}{\Delta t}$, where $N$ is the number of turns in the coil, $\Delta \Phi_B$ is the change in magnetic flux, and $\Delta t$ is the time over which this change occurs.The magnetic flux ($\Phi_B$) through the coil is given by $\Phi_B = B \cdot A$, where $B$ is the magnetic field and $A$ is the area of the coil. The area of the square coil is $A = (0.18 m)^2 = 0.0324 m^2$.The change in magnetic flux ($\Delta \Phi_B$) is then $\Delta B \cdot A = (0.50 T - 0.00 T) \cdot 0.0324 m^2 = 0.0162 T \cdot m^2$.Given that the change occurs over $\Delta t = 8.0 s$ and there are $N = 200$ turns, the induced emf is:$$\epsilon = -200 \cdot \frac{0.0162 T \cdot m^2}{8.0 s} = -200 \cdot 0.002025 V = -0.405 V$$The negative sign indicates the direction of the induced emf (opposing the change in flux), which is not needed for the magnitude of the emf. Therefore, the magnitude of the induced emf is $0.405 V$ or $0.41 V$, making the correct answer D) 0.41 V.

A Flat Coil Is Wrapped with 200 Turns of Very B⃗\vec { B }B

A Flat Coil Is Wrapped with 200 Turns of Very $\vec { B }$