Answer:
F = 0i (in the x-direction), 0j (in the y-direction),-8.59*10^-4 N k (In the z-direction)
Explanation:
The force given by charged particles moving in a magnetic field is given below (cross is cross product, they don't have that format in the equation tool):
[tex]F=qv (cross) B\\[/tex]
Now we can perform the cross product between v and B
[tex]v(cross)B = \left[\begin{array}{cc}4.19*10^{4} &-3.85*10^{4}\\1.8&0&\en[tex]v(cross)B = 69300 (kg*m/(s^2*C))\\[/tex]d{array}\right][/tex]
Now multiply by Q (charge) to get the force
[tex]F = -1.24*10^-8 * 69300\\F = -8.59*10^-4N[/tex]
F = -8.59*10^-4 N k
F = 0i, 0j, (-8.59*10^-4)k
The particle’s force is ( - 8.59 × 10⁻⁴ k ) N
[tex]\texttt{ }[/tex]
Let's recall magnetic force on moving charge as follows:
[tex]F = B q v \sin \theta[/tex]
where:
F = magnetic force ( N )
B = magnetic field strength ( T )
q = charge of object ( C )
v = speed of object ( m/s )
θ = angle between velocity and direction of the magnetic field
Let's tackle the problem now !
[tex]\texttt{ }[/tex]
Given:
speed of particle = v_y = -3.85 × 10⁴ m/s
magnetic field strength = B_x = 1.80 T
charge of particle = q = -1.24 × 10⁻⁸ C
direction of speed = θ = 90°
Asked:
the particle's force = F = ?
Solution:
[tex]F = B_x q v_y \sin \theta[/tex]
[tex]F = 1.80 \times 1.24 \times 10^{-8} \times 3.85 \times 10^4 \times \sin 90^o[/tex]
[tex]\boxed{F \approx 8.59 \times 10^{-4} \texttt{ N}}[/tex]
According to the Left Hand Rule , the direction of force is in negative z-axis.
We could write this force in vector form as follows:
[tex]\overrightarrow{F} = - (8.59 \times 10^{-4}) ~ \widehat{k} \texttt{ N}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Magnetic Field
