Chapter 21, Problem 13P

To determine

The maximum induced EMF in the antenna.

Answer to Problem 13P

Solution:

The maximum induced EMF in the antenna is $0.65\text{mV}$ east or west.

Explanation of Solution

The horizontal component of the Earth’s magnetic field is effective in exerting a vertical force on the charged particle in the antenna.

The magnetic force on the positive charge in the antenna is to be directed upwards and gave a maximum magnitude.

The car should move towards the East through the northward horizontal component of the magnetic field.

Given:

The angle of dip, that is, the downward angle $38°$.

The velocity of the 30.0 m/s

The length of antenna 55 cm.

The horizontal component of Earth’s magnetic field $5.0\times {10}^{-5}\text{ T}$.

Formula Used:

The expression for the magnetic force is as follows:

$F=qvB\sin \theta$

Here, $\theta$ is the angle between the velocity vector and the magnetic field vector, q is the charge, v is the velocity vector, and B is the magnetic field.

The maximum induced EMF in the antenna is as follows:

$\epsilon =Blv$

Here, $\epsilon$ is the induced EMF, l is the length, and B is the horizontal component of magnetic field.

Calculation:

Now, substitute $B\cos \theta$ for B, 55 cm for l, and 30.0 m/s for v in the equation $\epsilon =Blv$.

$\epsilon =B\cos \theta \left(55\text{ cm}\right)\left(30.0\text{ m/s}\right)$

Substitute $38°$ for $\theta$ and $5.0\times {10}^{-5}\text{ T}$ for B in the equation $\epsilon =B\cos \theta \left(55\text{ cm}\right)\left(30.0\text{ m/s}\right)$.

$\begin{array}{c}\epsilon =5.0\times {10}^{-5}\text{ T}\cos 38°\left(55\text{ cm}\right)\left(30.0\text{ m/s}\right)\\ =\left(5.0\times {10}^{-5}\text{ T}\right)\cos 38°\left(55\text{ cm}\right)\left(30.0\text{ m/s}\right)\\ =\left(5.0\times {10}^{-5}\text{ T}\right)\left(0.955\right)\left(55\text{ cm}\right)\left(\frac{{10}^{-2}\text{ m}}{1.00\text{ cm}}\right)\left(30.0\text{ m/s}\right)\\ =65.01\times {10}^{-5}\text{ V}\end{array}$