A 5-turn square loop (0.150 m along a side,resistance = 2.00 $\varOmega$)is placed in a magnetic field that makes an angle of 30 $\circ$ with the normal to the plane of the loop.The magnitude of this field varies with time according to B = 2.40t⁴,where t is measured in s and B in T.What is the induced current in the coil at t = 1.25 s? A)1.47 A B)1.23 A C)0.913 A D)0.527 A E)0.351 A

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The induced emf ($\epsilon$) in the coil 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 coil. The magnetic flux ($\Phi$) through a coil is given by $\Phi = NBA\cos(\theta)$, where $N$ is the number of turns, $B$ is the magnetic field, $A$ is the area of the coil, and $\theta$ is the angle between the magnetic field and the normal to the plane of the coil.Given that $B = 2.40t^4$ (T), $N = 5$, $A = 0.150^2$ m$^2$ (since it's a square loop), and $\theta = 30^\circ$, we can calculate the rate of change of flux ($\frac{d\Phi}{dt}$) and hence the induced emf ($\epsilon$) at $t = 1.25$ s. The resistance ($R$) of the loop is given as $2.00 \Omega$, which allows us to find the induced current ($I$) using Ohm's law ($I = \frac{\epsilon}{R}$).The calculation involves differentiating $B$ with respect to $t$ to find $\frac{dB}{dt}$, then calculating $\frac{d\Phi}{dt}$ by substituting $\frac{dB}{dt}$, $N$, $A$, and $\cos(\theta)$ into the formula for $\frac{d\Phi}{dt}$. Finally, using $\epsilon = -\frac{d\Phi}{dt}$ and $I = \frac{\epsilon}{R}$, we find the induced current at $t = 1.25$ s. The correct calculation leads to the answer that matches option C, which is approximately 0.913 A. This involves applying calculus and electromagnetic principles correctly to solve for the induced current.

A 5-Turn Square Loop (0 Ω\varOmegaΩ )Is Placed in a Magnetic Field That Makes an Angle

A 5-Turn Square Loop (0 $\varOmega$ )Is Placed in a Magnetic Field That Makes an Angle