Given \(\mathbf{H}=300 \mathbf{a}_{z} \cos \left(3 \times 10^{8} t-y\right) \mathrm{A} / \mathrm{m}\) in free space, find the emf developed in the general \(\mathbf{a}_{\phi}\) direction about the closed path having corners at \((a)(0,0,0),(1,0,0),(1,1,0)\), and \((0,1,0) ;(b)(0,0,0)(2 \pi, 0,0)\), \((2 \pi, 2 \pi, 0)\), and \((0,2 \pi, 0)\)

Faraday's Law of electromagnetic induction states that the emf (electromotive force) induced in a closed loop is equal to the negative rate of change of magnetic flux with time. Mathematically, it is represented as: \(\epsilon = -\frac{d\Phi}{dt}\) where \(\epsilon\) is the emf and \(\Phi\) is the magnetic flux. We start by finding the magnetic flux through the closed loop in both cases (a) and (b). #Step1a# Calculate the magnetic flux for the first closed path Consider the closed path with corners at \((0,0,0),(1,0,0),(1,1,0)\), and \((0,1,0)\). We can represent the area vector of the loop (A) as: \(\mathbf{A} = (1-0)(1-0)\mathbf{a}_{z} = \mathbf{a}_{z}\) The magnetic flux Φ through this closed loop is given by the dot product of the magnetic field vector \(\mathbf{H}\) and the area vector \(\mathbf{A}\): \(\Phi = \int_{S} \mathbf{H} \cdot \mathbf{A} \, dS\) \(\Phi = \int_{S} \left[ 300 \mathbf{a}_{z} \cos \left(3 \times 10^{8} t-y\right) \right] \cdot \mathbf{a}_{z} \, dS\) #Step1b# Calculate the magnetic flux for the second closed path Now consider the closed path with corners at \((0,0,0),(2\pi, 0,0)\), \((2\pi, 2\pi, 0)\), and \((0,2\pi, 0)\). The area vector of this loop (A) can be represented as: \(\mathbf{A} = (2\pi - 0)(2\pi - 0)\mathbf{a}_{z} = 4\pi^2\mathbf{a}_{z}\) The magnetic flux Φ through this closed loop is given by the dot product of the magnetic field vector \(\mathbf{H}\) and the area vector \(\mathbf{A}\): \(\Phi = \int_{S} \mathbf{H} \cdot \mathbf{A} \, dS\) \(\Phi = \int_{S} \left[ 300 \mathbf{a}_{z} \cos \left(3 \times 10^{8} t-y\right) \right] \cdot 4\pi^2\mathbf{a}_{z} \, dS\) #Step2# Differentiate magnetic flux with respect to time Now we need to find the rate of change of magnetic flux with time in both scenarios. For the first scenario (a): \(\frac{d \Phi}{dt}=\int_{S} \frac{\partial}{\partial t}\left[ 300 \cos \left(3 \times 10^{8} t-y\right) \right] \, dS\) For the second scenario (b): \(\frac{d \Phi}{dt}=\int_{S} \frac{\partial}{\partial t}\left[ 300 \cos \left(3 \times 10^{8} t-y\right) \times 4\pi^2 \right] \, dS\) #Step3# Calculate the emf for both cases Finally, calculate the emf for both cases `a` and `b` using the formula \(\epsilon = -\frac{d\Phi}{dt}\): For the first scenario (a): \(\epsilon_a = -\frac{d\Phi_a}{dt}\) For the second scenario (b): \(\epsilon_b = -\frac{d\Phi_b}{dt}\)