A proton moving at 3.90 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.20 10-13 N. What is

Question

A proton moving at 3.90 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.20 10-13 N. What is the angle between the proton’s velocity and the field

in progress 0
Khánh Gia 3 years 2021-08-25T17:56:55+00:00 1 Answers 9 views 0

Answers ( )

  1. Dulcie
    0
    2021-08-25T17:58:14+00:00
    Reply

    Answer:

    \theta=40^0

    Explanation:

    The magnitude of the magnetic force is

    F_m=evB\sin\theta

    To find the angle, we make \sin\theta subject of the formula

    \implies \sin\theta=\frac{F_m}{evB}=\frac{7.20\times 10^{-13}}{1.6\times 10^{-19}\times 3.90\times 10^6\times 1.80}

    \implies \sin\theta=0.641025641

    \therefore \theta=\sin^{-1}=39.8683^0\\\implies \theta\approxeq 40^0

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )