An electron follows a helical path in a uniform magnetic field given by:B =(20i^−50j^−30k^)mTAt time t = 0, the electron’s velocity is given

Question

An electron follows a helical path in a uniform magnetic field given by:B =(20i^−50j^−30k^)mTAt time t = 0, the electron’s velocity is given by:⃗v=(40i^−30j^+50k^)m/sa. What is the angle ϕ between v and B. The electron’s velocity changes with time. Do b. its speed c. the angled. What is the radius of the helical path?

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Ngọc Diệp 2 months 2021-08-01T04:34:02+00:00 1 Answers 5 views 0

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    2021-08-01T04:35:57+00:00

    Answer:

    a) 1.38°

    b) 7.53*10^11 m/s/s

    c) 6.52*10^-9m

    Explanation:

    a) to find the angle you can use the dot product between two vectors:

    \vec{v}\cdot\vec{B}=vBcos\theta\\\\\theta=cos^{1}(\frac{\vec{v}\cdot\vec{B}}{vB})

    v: velocity of the electron

    B: magnetic field

    By calculating the norm of the vectors and the dot product and by replacing you obtain:

    B=\sqrt{(20)^2+(50)^2+(30)^2}=61.64mT\\\\v=\sqrt{(40)^2+(30)^2+(50)^2}=70.71m/s\\\\\vec{v}\cdot\vec{B}=[(20)(40)+(50)(30)-(30)(50)]mTm/s=800mTm/s\\\\\theta=cos^{-1}(\frac{800*10^{-3}Tm/s}{(70.71m/s)(61.64*10^{-3}T)})=cos^{-1}(0.183)=1.38\°

    the angle between v and B vectors is 1.38°

    b) the change in the speed of the electron can be calculated by the change in the momentum in the following way:

    \frac{dp}{dt}=F_e=qvBsin\theta\\\\\frac{dp}{dt}=(1.6*10^{-19}C)(70.71m/s)(61.64*10^{-3}T)(sin(1.38\°))=6.85*10^{-19}N

    due to the mass of the electron is a constant you have:

    \frac{dp}{dt}=\frac{mdv}{dt}=6.85*10^{-19}N\\\\\frac{dv}{dt}=\frac{6.85*10^{-19}N}{9.1*10^{-31}kg}=7.53*10^{11}(m/s)/s

    the change in the speed is 7.53*10^{11}m/s/s

    c) the radius of the helical path is given by:

    r=\frac{m_ev}{qB}=\frac{(9.1*10^{-31}kg)(70.71m/s)}{(1.6*10^{-19}C)(61.64*10^{-3}T)}=6.52*10^{-9}m

    the radius is 6.52*10^{-9}m

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