A square filamentary loop of wire is \(25 \mathrm{~cm}\) on a side and has a resistance of \(125 \Omega\) per meter length. The loop lies in the \(z=0\) plane with its corners at \((0,0,0),(0.25,0,0),(0.25,0.25,0)\), and \((0,0.25,0)\) at \(t=0\). The loop is moving with a velocity \(v_{y}=50 \mathrm{~m} / \mathrm{s}\) in the field \(B_{z}=8 \cos (1.5 \times\) \(\left.10^{8} t-0.5 x\right) \mu \mathrm{T}\). Develop a function of time that expresses the ohmic power being delivered to the loop.

To calculate the magnetic flux through the loop, we need to integrate the magnetic field over the loop's area: \(\Phi(t) = \int B_z \; dA\) Since the field is only in the z-direction and the loop is in the \(z=0\) plane, the angle between the magnetic field and the normal vector to the loop's area is 0. Thus, the integration simplifies to: \(\Phi(t) = \int \int B_{z} \; dxdy\) Now let's find the limits of integration. At any given time t, the position of the loop in the y-direction is given by \(y_{min} = 50t\) and \(y_{max} = 0.25 + 50t\), keeping in mind the side length of the loop is \(0.25 \mathrm{m}\). The x-direction doesn't change during the motion, so the limits are between 0 and 0.25 m. Taking these limits into account, we can rewrite the integral as follows: \(\Phi(t) = \int_{0}^{0.25} \int_{50t}^{0.25+50t} B_{z} \; dydx\) Now, insert the given magnetic field equation: \(\Phi(t) = \int_{0}^{0.25} \int_{50t}^{0.25+50t} 8 \cos \left(1.5 \times 10^8 t - 0.5 x\right) \mu T \; dydx\)

To find the rate of change of magnetic flux, take the derivative of \(\Phi(t)\) with respect to time: \(\dfrac{d\Phi}{dt} = \dfrac{d}{dt} \left[ \int_{0}^{0.25} \int_{50t}^{0.25+50t} 8 \cos \left(1.5 \times 10^8 t - 0.5 x\right) \mu T \; dydx \right]\)

The induced electromotive force (EMF) can be calculated using Faraday's law: \(EMF = -\dfrac{d\Phi}{dt}\)

Using Ohm's law to find the current flowing through the loop, we have: \(I(t) = \dfrac{EMF}{R}\) Here, R is the total resistance of the loop, which can be calculated as follows: \(R = \) resistance per meter \(\times\) total length of the loop \(= 125 \Omega/m \times \left(4 \times 0.25 \;m\right) = 125 \Omega\) Substituting the value of resistance: \(I(t) = \dfrac{EMF}{125 \Omega}\)

The ohmic power can be calculated using the power formula: \(P(t) = I^2(t) \times R\) Substitute the values of current and resistance: \(P(t) = \left( \dfrac{EMF}{125 \Omega} \right)^2 \times 125 \Omega\) Now the ohmic power being delivered to the loop is given by the above formula as a function of time.