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

Answer 2

a. The value of vₓ = 33.77 km/s

b. The value of Vy = 81.05 km/s

The magnetic force on the proton

The magnetic force, F on the proton moving in the uniform magnetic field is given by

F = qv × B where

• F = force on proton = (3.82 × 10⁻¹⁷ N)l + (1.59 × 10⁻¹⁷ N)j + 0k,
• q = charge on proton = 1.602 × 10⁻¹⁹ C,
• v = velocity of proton = Vxl + Vyj + (2000 m/s)k and
• B = magnetic field = (10i - 24.0j + 30k) mT =  (10i - 24.0j + 30k) × 10⁻³ T

Re-writing the force in matrix form, we have

$\left[\begin{array}{ccc}F_{x} &F_{y} &F_{z} \\\end{array}\right] = q\left[\begin{array}{ccc}i&j&k\\v_{x} &v_{y}&v_{z}\\B_{x}&B_{y}&B_{z}\end{array}\right]$

Taking the determinant, we have

$F_{x}i + F_{y}j + F_{z}k = q[(v_{y}B_{z} - v_{z}B_{y})]i + q[(v_{x}B_{z} - v_{z}B_{x})]j + q[(v_{x}B_{y} - v_{y}B_{x})]k$

Equating the components of the force, we have

$F_{x} = q[(v_{y}B_{z} - v_{z}B_{y})] (1)\\F_{y} = q[(v_{x}B_{z} - v_{z}B_{x})] (2)\\F_{z} = q[(v_{x}B_{y} - v_{y}B_{x})] (3)$

$F_{x}/q = [(v_{y}B_{z} - v_{z}B_{y})] (4)\\F_{y}/q = [(v_{x}B_{z} - v_{z}B_{x})] (5)\\F_{z}/q = [(v_{x}B_{y} - v_{y}B_{x})] (6)$

Since F = (3.82 × 10⁻¹⁷ N)l + (1.59 × 10⁻¹⁷ N)j + 0k, equation (5) and (6) become

$1.59 X 10^{-17} /1.602 X 10^{-19} = [(v_{x}B_{z} - v_{z}B_{x})] (5)\\993 = [(v_{x}B_{z} - v_{z}B_{x})] (7)\\Also\\0/q = [(v_{x}B_{y} - v_{y}B_{x})] (6)\\0 = (v_{x}B_{y} - v_{y}B_{x}) \\v_{x}B_{y} = v_{y}B_{x}\\v_{y} = \frac{v_{x}B_{y}}{B_{x}} (8)$

a. The value of Vx

The value of vₓ = 33.77 km/s

Since $B_{x}$ = 10 × 10 ⁻³ T, $B_{z}$ = 30 × 10 ⁻³ T and $v_{z}$ = 2000 m/s, substituting the values of the variables into equation (7), we have

$v_{x}B_{z} - v_{z}B_{x} = 993 (7)$

vₓ(30 × 10 ⁻³ T) - 2000 m/s × 10 × 10 ⁻³ T = 993 N

vₓ(30 × 10 ⁻³ T) - 20 m/sT = 993 N

vₓ(30 × 10 ⁻³ T) = 993 N + 20 m/sT

vₓ(30 × 10 ⁻³ T) = 1013 N

vₓ = 1013 N/(30 × 10 ⁻³ T)

vₓ = 33.77 × 10³ m/s

vₓ = 33.77 km/s

So, the value of vₓ = 33.77 km/s

b. The value of Vy

The value of Vy = 81.05 km/s

Since $B_{x}$ = 10 × 10 ⁻³ T, $B_{y}$ = 24.0 × 10 ⁻³ T and $v_{x}$ = 33.77 × 10³ m/s, substituting the values of the variables into equation (8), we have

$v_{y} = \frac{v_{x}B_{y}}{B_{x}} (8)$

Vy = 33.77 × 10³ m/s × 24.0 × 10 ⁻³ T/10 × 10 ⁻³ T

Vy = 810.48 m/sT/10 × 10 ⁻³ T

Vy = 81.048 × 10³  m/s

Vy ≅ 81.05 km/s

So, the value of Vy = 81.05 km/s

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