The external magnetic field in a loop decreases from 0.1 Tesla to zero in 2 seconds.The loop of wire has ten turns,a radius of 10 cm,with its axis 45 degrees relative to the magnetic field.How large is the induced voltage in the loop?
A)0 mV
B)11 mV
C)9.4 mV
D)32.5 mV
E)0.31 mV

Question

Quizplus · Accepted Answer

The induced voltage (emf) in a loop can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf is equal to the negative rate of change of the magnetic flux through the loop. The magnetic flux ($\Phi$) is given by $\Phi = BA\cos(\theta)$, where $B$ is the magnetic field strength, $A$ is the area of the loop, and $\theta$ is the angle between the magnetic field and the normal to the plane of the loop.Given:- Initial magnetic field ($B_i$) = 0.1 Tesla- Final magnetic field ($B_f$) = 0 Tesla- Change in time ($\Delta t$) = 2 seconds- Number of turns ($N$) = 10- Radius of the loop ($r$) = 10 cm = 0.1 m (convert cm to m)- Angle ($\theta$) = 45 degreesFirst, calculate the area of the loop ($A$) = $\pi r^2$ = $\pi (0.1)^2$ m$^2$.The change in magnetic flux ($\Delta \Phi$) = $N \Delta (BA\cos(\theta))$ = $N(B_f - B_i)A\cos(\theta)$.Substituting the given values:- $A = \pi (0.1)^2$- $\cos(45^\circ) = \frac{\sqrt{2}}{2}$$\Delta \Phi = 10 \times (0 - 0.1) \times \pi (0.1)^2 \times \frac{\sqrt{2}}{2}$The induced emf ($emf$) = $-\Delta \Phi / \Delta t$.Substituting the values and calculating gives an induced emf of approximately 11 mV, making option B the correct answer.

The External Magnetic Field in a Loop Decreases from 0