Question

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50km/s in the +x-direction experiences a force of 2.06�10-16N in the +y-direction, and an electron moving at 4.40km/s in the -z-direction experiences a force of 8.40�10-16N in the +y-direction.
Part A

What is the magnitude of the magnetic field?

B = T
Part B

What is the direction of the magnetic field? (in the xz-plane)

theta = from the -z-direction
Part C

What is the magnitude of the magnetic force on an electron moving in the -y-direction at 3.70km/s ?

F = N
Part D

What is the direction of this the magnetic force? (in the xz-plane)

theta = from the -x-direction

Answer 1

Answer:

Explanation:

velocity of proton  v = 1.5 x 10³ i  m /s

charge on proton e = 1.6 x 10⁻¹⁹ C

Let the magnetic field be B = Bx i + Bz k

force on charged particle ( proton )

F = e ( v x B )

2.06  x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ [ 1.5 x 10³ i x ( Bx i + Bz k) ]

2.06  x10⁻¹⁶ j = - 1.6 x 10⁻¹⁹ x 1.5 x 10³  Bz j) ]

2.06  x10⁻¹⁶ = - 1.6 x 10⁻¹⁹ x 1.5 x 10³  Bz

Bz = -  .8583  

force on charged particle ( electron )

F = e ( v x B )

8.40  x10⁻¹⁶ j = -1.6 x 10⁻¹⁹ [ - 4.4 x 10³ k  x ( Bx i + Bz k) ]

8.4  x10⁻¹⁶ j =  1.6 x 10⁻¹⁹ x 4.4 x 10³  Bx j ]

- 8.4  x10⁻¹⁶ =  1.6 x 10⁻¹⁹ x 4.4 x 10³  Bx

Bx =  - 1.19

Magnetic field = - 1.19 i - .8583 k

magnitude = √ (1.19² + .8583²)

= 1.467 T

If it is making angle θ with x - axis in x -z plane

Tanθ = (.8583 / 1.19 )

36⁰ .

C )

v = - 3.7 x 10³j m /s

e = - 1.6 x 10⁻¹⁶ C

Force = F = e ( v x B )

= -1.6 x 10⁻¹⁹ [ -3.7 x 10³ j  x ( Bx i + Bz k) ]

=  - 1.6 x 10⁻¹⁹ x 3.7 x 10³  Bx  k -1.6 x 10⁻¹⁹ x 3.7 x 10³Bzi ]

=    5.08 i - 7.04 k

Tanθ = 54 ° .