Question

A proton moves through a uniform magnetic field given by vector B = (10 i hat- 24.0 j hat + 30 k hat) mT. At time t1, the proton has a velocity given by vector v = v x i hat + vy j hat + (2000 m/s) k hat and the magnetic force on the proton is vector Fb = (3.82x10-17 N) i hat + (1.59x10-17 N) j hat. (a) At that instant, what is vx? m/s (b) What is vy?

Answer 1

Answer:

a)vₓ = 666.67 m / s, b)  $v_{y}$ = -93.3 m / s

Explanation:

The expression for the magnetic force is

    F = q v x B

Bold indicate vectors, the way to solve this problem is to use determinants

$\left[\begin{array}{ccc}F_x&F_y&F_z\end{array}\right]$  = $\left[\begin{array}{ccc}i&j&k\\v_x&v_y&v_z\\B_x&B_y&B_z\end{array}\right]$

The given system has the vectors

    F = (3.82 i ^ + 1.59 j ^ + 0 K`) 10⁻¹⁷ N

    v = (vₓ i ^ + $v_{y}$ i ^ + 2000) m / s

    B = (10 i ^ - 24 j ^ + 30 k ^) 10⁻³ T

Let's write and solve the determinant

$\left[\begin{array}{ccc}3.82&1.59&0\end{array}\right]$ 10⁻¹⁷ = $\left[\begin{array}{ccc}i&j&k\\v_x&v_y&2000\\10&-24&30\end{array}\right]$ 10⁻³

X axis

    3.82 10⁻¹⁴ = $v_{y}$ 30 - 2000 (-24)

    30 $v_{y}$ = 3.82 10⁻¹⁴ - 2800

    $v_{y}$ = -2800/30

    $v_{y}$ = -93.3 m / s

Y Axis  

    1.59 10⁻¹⁴ = 2000 10 -vₓ 30

    30 vₓ = 20000 + 1.59 10⁻¹⁴

    vₓ = 20000/30

    vₓ = 666.67 m / s