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Chapter 22: Problem 10

Using Faraday's law, show that if a magnetic dipole antenna's axis makes an angle \(\theta\) with the magnetic field of an EM wave, the induced emf in the antenna is reduced from its maximum possible value by a factor of cos $\theta .$ [Hint: Assume that, at any instant, the magnetic field everywhere inside the loop is uniform. \(]\)

Short Answer

Expert verified
In this problem, we applied Faraday's law to show that the induced emf in a magnetic dipole antenna is reduced from its maximum possible value by a factor of cos(θ) when the antenna's axis makes an angle θ with the magnetic field of an EM wave. We started by stating Faraday's law and then calculated the magnetic flux through the loop. Applying Faraday's law, we arrived at the expression for the induced emf, considering the magnetic field and its time-dependence. After differentiating the magnetic flux with respect to time, we simplified the expression to arrive at the final result. We showed that the induced emf is reduced by a factor of cos(θ) compared to its maximum value when the angle between the antenna's axis and the magnetic field is θ.

Step by step solution

01

State Faraday's law

Faraday's law states that the induced electromotive force (emf) in any closed circuit is equal to the rate at which the magnetic flux through the circuit changes with time. Mathematically, it can be written as: \(\text{emf} = -\dfrac{d\Phi_B}{dt}\) where \(\Phi_B\) is the magnetic flux and \(t\) is the time.
02

Calculate the magnetic flux

Now, let's calculate the magnetic flux through the loop. Consider a loop with area A and a magnetic field B exactly perpendicular to its plane. If the loop makes an angle θ with the magnetic field, then magnetic flux is given by: \(\Phi_B = B \cdot A \cdot \cos{\theta}\) where B is the magnitude of the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop's plane.
03

Apply Faraday's law to get the induced emf

Now, let's apply Faraday's law to calculate the induced emf. We know emf is given by \(-\dfrac{d\Phi_B}{dt}\), so we have: \(\text{emf} = -\dfrac{d(B \cdot A \cdot \cos{\theta})}{dt}\) Now, let \(B = B_0 \cos{(\omega t)}\), where \(B_0\) is the maximum value of the magnetic field and \(\omega\) is the angular frequency of the EM wave. Substituting the magnetic field expression in the above equation, we get: \(\text{emf} = -\dfrac{d(B_0 \cos{(\omega t)} \cdot A \cdot \cos{\theta})}{dt}\)
04

Differentiate the magnetic flux with respect to time

Now, we need to differentiate the expression for the magnetic flux with respect to time to find the emf. As θ and A do not change with time, we can treat them as constants and differentiate only the time-dependent part of the expression. We get: \(\text{emf} = -A \cdot \cos{\theta} \cdot B_0 \cdot \dfrac{d(\cos{(\omega t)})}{dt} = -A \cdot \cos{\theta} \cdot B_0 \cdot (-\omega \sin{(\omega t)})\)
05

Simplify to get the final expression for the induced emf

By simplifying the right-hand side of the equation, we get the final expression for the induced emf: \(\text{emf}(\theta, t) = A \cdot \cos{\theta} \cdot B_0 \cdot \omega \sin{(\omega t)}\) Now, let's look at the maximum possible value of the induced emf. This occurs when the angle θ = 0, meaning that the magnetic field and the antenna's axis are parallel to each other. For this case, the induced emf is given by: \(\text{emf}_{\text{max}}(t) = A \cdot B_0 \cdot \omega \sin{(\omega t)}\) By comparing the expressions for the induced emf in both cases, it can be seen that the induced emf is reduced from its maximum possible value by a factor of cos(θ): \(\text{emf}(\theta, t) = \cos{\theta} \cdot \text{emf}_{\text{max}}(t)\)

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