Question

A particle with a charge of −1.24×10−8c is moving with instantaneous velocity v⃗ = (4.19×104m/s)i^ + (−3.85×104m/s)j^ . part a what is the force exerted on this particle by a magnetic field b⃗ = (1.80 t ) i^? enter the x, y, and z components of the force separated by commas.

Answer 1

The particle’s force is ( - 8.59 × 10⁻⁴ k ) N

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Further explanation

Let's recall magnetic force on moving charge as follows:

$F = B q v \sin \theta$

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 !

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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:

$F = B_x q v_y \sin \theta$

$F = 1.80 \times 1.24 \times 10^{-8} \times 3.85 \times 10^4 \times \sin 90^o$

$\boxed{F \approx 8.59 \times 10^{-4} \texttt{ N}}$

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:

$\overrightarrow{F} = - (8.59 \times 10^{-4}) ~ \widehat{k} \texttt{ N}$

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Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Magnetic Field

Answer 2

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):

$F=qv (cross) B$

Now we can perform the cross product between v and B

$v(cross)B = \left[\begin{array}{cc}4.19*10^{4} &-3.85*10^{4}\\1.8&0&\en[tex]v(cross)B = 69300 (kg*m/(s^2*C))$d{array}\right][/tex]

Now multiply by Q (charge) to get the force

$F = -1.24*10^-8 * 69300\\F = -8.59*10^-4N$

F = -8.59*10^-4 N k

F = 0i, 0j, (-8.59*10^-4)k