An electron that has a velocity with x component 2.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic field with x c

Question

An electron that has a velocity with x component 2.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic field with x component 0.029 T and y component -0.14 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.

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Trung Dũng 1 month 2021-08-06T14:43:03+00:00 1 Answers 6 views 0

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    2021-08-06T14:44:23+00:00

    Answer with Explanation:

    We are given that

    v_x=2.6\times 10^6 m/s

    v_y=2.4\times 10^6 m/s

    B_x=0.029 T

    B_y=-0.14 T

    a.We have to find the magnitude of the magnetic force on the electron.

    v\times B=\begin{vmatrix}i&j&k\\2.6\times 10^6&2.4\times 10^6&0\\0.029&-0.14&0\end{vmatrix}

    v\times B=(-0.364-0.0696)\times 10^6 k=-0.4336\times 10^6 k

    Charge on an electron,q=-1.6\times 10^{-19} C

    F=q(v\times B)=\mid -1.6\times 10^{-19}(-0.4336)\times 10^6\mid =6.9\times 10^{-14} N

    Force act along positive z- direction.

    b.Charge on proton=q=1.6\times 10^{-19} C

    F=\mid 1.6\times 10^{-19}(-0.4336)\times 10^6\mid =6.9\times 10^{-14} N

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