Question

32.2 induced magnetic fields
a small cart (which you may treat as a point charge) carrying a charge of 2.000 c moves along a straight track at a constant speed of 8.000 m/s. what is the magnitude of the magnetic field at a point beside the track that is 9.000 cm away from the track and 27.000 cm farther down the track than the charged cart? μt
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Explanation:

Step1: Calculate the distance \( r \) from the charge

Use the Pythagorean theorem \( r=\sqrt{x^{2}+y^{2}} \), where \( x = 9.000\ cm=0.09\ m \) and \( y = 27.000\ cm = 0.27\ m \).

$$ r=\sqrt{(0.09)^{2}+(0.27)^{2}}=\sqrt{0.0081 + 0.0729}=\sqrt{0.081}\ m $$

Step2: Use the formula for magnetic field due to a moving charge

The formula for the magnetic field due to a moving charge is \( B=\frac{\mu_{0}}{4\pi}\frac{qv\sin\theta}{r^{2}} \). Since the velocity \( v \) and the position vector \( \vec{r} \) are perpendicular (\(\sin\theta = 1\), \(\frac{\mu_{0}}{4\pi}=10^{- 7}\ T\cdot m/A\)), \( q = 2.000\ C\), \( v=8.000\ m/s \).

$$ B = 10^{-7}\times\frac{2\times8}{0.081} $$

$$ B=\frac{16\times10^{-7}}{0.081}\ T $$

$$ B=\frac{16\times10^{-7}}{0.081}\times10^{6}\ \mu T $$

$$ B=\frac{16\times10^{-1}}{0.081}\ \mu T $$

$$ B=\frac{1.6}{0.081}\ \mu T\approx19.88\ \mu T $$

Answer:

\(19.9\ \mu T\)