**Educational Content: Magnetic Field of a Moving Electron****Diagram Explanation:**The diagram shows an electron moving with a velocity \( \vec{v} \) at \( 3.60 \times 10^7 \, \text{m/s} \). The electron is moving in the negative y-direction. The coordinate axes are labeled as \( x \) and \( y \), with the electron positioned at the origin. There's a point \( P \) located 6.00 micrometers away from the electron at a 35-degree angle below the x-axis.**Problem Statement:**An electron moves at \( 3.60 \times 10^7 \, \text{m/s} \) as shown in the figure. Find the magnitude and direction of the magnetic field this electron produces at point \( P \), which is \( 6.00 \, \mu \text{m} \) away from the electron. \((e = 1.60 \times 10^{-19} \, \text{C})\).**Answer Options:**- \( \circ \) \( 9.18 \times 10^{-9} \, \text{T, out of the page} \)- \( \circ \) \( 1.53 \times 10^{-3} \, \text{T, out of the page} \)- \( \circ \) \( 1.60 \times 10^{-8} \, \text{T, into the page} \)- \( \circ \) \( 2.18 \times 10^{-3} \, \text{T, out of the page} \)- \( \circ \) \( 1.31 \times 10^{-8} \, \text{T, into the page} \)- \( \circ \) \( 1.53 \times 10^{-3} \, \text{T, into the page} \)- \( \circ \) \( 9.18 \times 10^{-9} \, \text{T, into the page} \)- \( \circ \) \( 2.18 \times 10^{-3} \, \text{T, into the page} \)- \( \circ \) \( 1.60 \times 10^{-8} \, \text{T,

Answered: Image /qna-images/answer/b5646c50-f0d0-4e47-9226-4c61cfae02bf.jpg

Answered: P 35% Electron 9.18 x 107 T. out of the…

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