Question

An electron moving with a speed of 6.2×105 m/s in the positive x direction experiences zero magnetic force. When it moves in the positive y direction, it experiences a force of 3.9×10−13 N that points in the positive z direction. What is the direction of the magnetic field? What is the magnitude of the magnetic field?

Answer #1

A -4.60 μC charge is moving at a constant speed of
6.70×105 m/s in the +x−direction relative to a reference
frame. At the instant when the point charge is at the origin, what
is the magnetic-field vector it produces at the following
points.
C) x=0.500m, y=0.500m, z=0

A beam of protons is moving in the +x direction with a
speed of 13 km/s through a region in which the electric field is
perpendicular to the magnetic field. The beam of protons is not
deflected in this region. The magnetic field has a magnitude of 0.8
T and points in the +y direction. Therefore, the magnitude
of the electric field is 11.570 V/m (= (13 x 10³ m/s) (0.8T)) and
the direction is along the negative z-axis.
What...

1. if electron moving in the +x direction and a force in the -z
direction , what is the magnetic field direction?
2. a negative charge (electron) moves in the +z direction and no
magnetic force due to the magnetic field, then what is the
direction of magnetic field?
what is the dirction of magnetic field, if there is magnetic
force?

A -4.80 μC charge is moving at a constant speed of
6.80×105 m/s in the +x−direction relative to a reference
frame. At the instant when the point charge is at the origin, what
is the magnetic-field vector it produces at the following
points.
Part A
x=0.500m,y=0, z=0
Enter your answers numerically separated by commas.
Part B
x=0, y=0.500m, z=0
Part C
x=0.500m, y=0.500m, z=0
Enter your answers numerically separated by commas.
Part D
x=0, y=0, z=0.500m

When at rest, a proton experiences a net electromagnetic force
of magnitude 8.5×10−13 N pointing in the positive x direction. When
the proton moves with a speed of 1.4×106 m/s in the positive y
direction, the net electromagnetic force on it decreases in
magnitude to 7.5×10−13 N , still pointing in the positive x
direction.
PART A
Find the
magnitude of the electric field.
PART
B
Find the direction of the electric field.
a. positive x direction
b. negative...

When at rest, a proton experiences a net electromagnetic force
of magnitude 9.0×10?13 Npointing in the positive
x direction. When the proton moves with a speed of
1.4×106 m/s in the positive y direction, the
net electromagnetic force on it decreases in magnitude to
7.5×10?13 N , still pointing in the positive x
direction.
Part A:Find the magnitude of the electric field.
Part B: Find the direction of the electric field
Part C: Find the magnitude of the magnetic field....

When at rest, a proton experiences a net electromagnetic force
of magnitude 8.1×10−13 N pointing in the positive
x direction. When the proton moves with a speed of
1.6×106 m/s in the positive y direction, the
net electromagnetic force on it decreases in magnitude to
7.0×10−13 N , still pointing in the positive x
direction.
Part A: Find the magnitude of the electric
field. Express your answer using two significant figures.
Part B: Find the direction of the electric
field....

a) An electron has a velocity of 7.0 x 106 m/s in the
positive x direction at a point where the magnetic field has the
components Bx = 3.0 T, By = 2.5 T, and Bz =
2.0 T. What is the magnitude of the acceleration of the electron at
this point (in 1018 m/s2)? (me = 9.11 x
10-31 kg, e = 1.6 x 10-19 C)
b) A particle (q = 5.0 nC, m = 3.0 μg) moves in...

An electron with a velocity given by v⃗ =(1.6×105 m/s
)x^+(6700 m/s )y^ moves through a region of space with a magnetic
field B⃗ ==(0.26 T )x^−(0.10 T )z^ and an electric field E⃗ =(220
N/C )x^.
Using cross products, find the magnitude of the net force acting
on the electron. (Cross products are discussed in Appendix A.)
Express your answer using two significant figures.

An electron experiences a magnetic force of magnitude 4.60 x
10-15 N when moving at an angle of 60.0° with respect to
a magnetic field of magnitude 3.50 ´ 10-3 T. Indicate
respective directions of magnetic field vector and the force
vector. Find the velocity of the electron (its magnitude and
direction).

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