Question

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.

Short answer

Answer: The ohmic power being delivered to the loop as a function of time is given by the formula: \(P(t) = \left( \dfrac{EMF}{125 \Omega} \right)^2 \times 125 \Omega\), where EMF is calculated using Faraday's law and has a time-dependent expression.

Step-by-step solution

Calculate the magnetic flux through 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: \(\Phi(t) = \int_{0}^{0.25} \int_{50t}^{0.25+50t} 8 \cos \left(1.5 \times 10^8 t - 0.5 x\right) \mu T \; dydx\)

Determine the rate of change of magnetic flux

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

Calculate the induced EMF

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

Obtain the current in the loop

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: \(I(t) = \dfrac{EMF}{125 \Omega}\)

Calculate the ohmic power being delivered to the loop

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.