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

Step-by-step 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 ° .