Magnetic fields are produced by moving charges and exert
forces on moving charges. When a particle with charge q is moving
with velocity v
S
in a magnetic field B
S
, the force F
S
that the field exerts
on the particle is given by F
S
= qv
S
: B
S
. The SI units are as follows: For
charge it is the coulomb (C), for magnetic field it is tesla (T), for force
it is newton (N), and for velocity it is m>s. If q = -8.00 * 10-6 C, v
S
is 3.00 * 104 m>s in the +x-direction, and B
S
is 5.00 T in the -y-direction, what are the magnitude and direction of the force that the magnetic
field exerts on the charged particle?
F = q V cross B
q = -8 * 10^-6
V = 3 * 10^4 in x direction
B = 5 in - y direction
V cross B is in NEGATIVE z direction by right hand rule with positive x cross negative y
since q is also negative, the final result is in the PLUS z direction
so
+ 8*10^-6 * 3*10^4 * 5 Newtons
The magnitude of the force that the magnetic field exerts on the charged particle can be calculated using the formula F = qvB, where q is the charge, v is the velocity, and B is the magnetic field strength. Plugging in the given values: q = -8.00 * 10^-6 C, v = 3.00 * 10^4 m/s, and B = 5.00 T, we get:
F = (-8.00 * 10^-6 C) * (3.00 * 10^4 m/s) * (5.00 T)
Now, let me grab my calculator... *clown noises*
Calculating... calculating... Oops, slipped on a banana peel!
Okay, let me try again... *clown noises*
Calculating...
Alright, the result is -0.060 N.
Now, for the direction of the force, we can determine it using the right-hand rule. If you point your right-hand thumb in the direction of the velocity (in the +x-direction), and curl your fingers towards the magnetic field direction (in the -y-direction), your palm will be facing the direction of the force.
So, in this case, the force is in the -z-direction, which means it is pointing downward.
To find the magnitude and direction of the force exerted by the magnetic field on the charged particle, follow these steps:
Step 1: Given values
- Charge of the particle, q = -8.00 x 10^(-6) C
- Velocity of the particle, vS = 3.00 x 10^4 m/s in the +x-direction
- Magnetic field, BS = 5.00 T in the -y-direction
Step 2: Use the formula for calculating the force exerted by a magnetic field, F = qvB.
Step 3: Substitute the given values into the formula:
F = (-8.00 x 10^(-6) C) x (3.00 x 10^4 m/s) x (5.00 T)
= -1.20 x 10^(-1) N
Step 4: The magnitude of the force is given by the absolute value of -1.20 x 10^(-1) N:
|F| = 1.20 x 10^(-1) N
Step 5: Since the charge is negative and the magnetic field is in the -y-direction, the force will be in the +z-direction.
Therefore, the magnitude of the force that the magnetic field exerts on the charged particle is 1.20 x 10^(-1) N, and the direction is in the +z-direction.
To find the magnitude and direction of the force that the magnetic field exerts on the charged particle, you can use the formula:
F = q × v × B × sin(θ)
where:
F is the force,
q is the charge of the particle,
v is the velocity of the particle,
B is the magnetic field strength,
and θ is the angle between the velocity vector and the magnetic field vector.
In this case, q = -8.00 * 10^-6 C, v = 3.00 * 10^4 m/s in the +x-direction, and B = 5.00 T in the -y-direction.
First, let's find the angle θ between the velocity vector and the magnetic field vector. Since the velocity is in the +x-direction and the magnetic field is in the -y-direction, the angle θ can be calculated as:
θ = 90 degrees - 0 degrees
θ = 90 degrees
Now we can plug in the values into the formula and calculate the force:
F = (-8.00 * 10^-6 C) × (3.00 * 10^4 m/s) × (5.00 T) × sin(90 degrees)
Since sin(90 degrees) = 1, the calculation becomes:
F = (-8.00 * 10^-6 C) × (3.00 * 10^4 m/s) × (5.00 T) × 1
F = -1.20 N
The magnitude of the force is 1.20 N. The negative sign indicates that the force is in the opposite direction of the velocity vector.
Therefore, the magnitude of the force is 1.20 N, and the direction is opposite to the direction of the velocity vector.