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The velocity is along the z axis, which is perpendicular to the x-y plane, which means that the angle between v and B is 90 degrees. So the force will be F = qvB = (3*10^-6)(10)(0.5)(sin 90) = 15*10^-6 N at 90 degrees.For part b, the direction of v changes to the z-axis, so the angle between v and B will also change. To find the new angle, we can use the trigonometric identity sin(A + B) = sin(A)cos(B) + cos(A)sin(B). Since we know the angle between v and B in part a (41.83 degrees
- #1
scholio
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Homework Statement
a 3 microcoulomb charge having a speed of 10m/s is fired into a region of constant magnetic field where B is given by:
B = 0.3i - 0.4j
a) the magnitude force on the charge is found to be 2*10^-6 Newtons. what is the angle between v and B?
b) if the direction of v is changed so that the charge moves along the +z direction at a speed of 10m/s, what would be the magnitude and direction of the magnetic force in this case?i am supposed to get 7.66 degrees for part a, and 15*10^-6 Newtons at 36.9 degrees for part b
Homework Equations
radius r = mv/qB where m is mass, v is velocity, q is charge, B is magnetic field
electromagnetic force, F = qE + qv X B where X indicates cross product, E is electric field
The Attempt at a Solution
part a:
using F = qE + qv X B ----> find theta
F = qE + qv X B
2*10^-6 = (3*10^-6E) + ((3*10^-6)(10)(0.3i - 0.4j)sin(theta))
sin(theta) = 2*10^-6 - 3*10^-6E/((3*10^-6)(10)(0.3i - 0.4j))
theta = inverse sin((2*10^-6 - 3*10^-6E)/(9*10^-6i - 1.2*10^-5j)) holding electric field E constant
is this the correct approach to find theta?
i haven't started part b yet, but i am assuming theta changes to (theta from part a + 90 degrees) and i just substitute that new theta into the electromagnetic force equation to solve for force.
is my approach for part a, correct? how about part b? or am i using the wrong equation?
i am supposed to get 7.66 degrees for part a, and 15*10^-6 Newtons at 36.9 degrees for part b
any help appreciated
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- #2
fizikx
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hint: the magitude of v is already given, you have to calculate the magnitude of B then find theta by:
F = q(v X B) = q|v||B|sin(theta)
hope this helps
fizikx
- #3
dynamicsolo
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While your general statement of the force on an electric charge is correct, there is no electric field implied in this problem. You can just set E = 0.
- #4
scholio
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oh you a right, that equation was right in front of me, must've forgotten.
F = qvB sin(theta) --> find theta
2*10^-6 = (3*10^-6)(10)(0.3i -0.4j) sin(theta)
sin(theta) = (2*10^-6)/(9*10^-6i - 1.2*10^-5j)
theta = inverse sine [(2*10^-6)/(9*10^-6i - 1.2*10^-5j)]
theta = inverse sine[(2*10^-6)/(-3*10^-6)]
theta = inverse sine(-0.667)
theta = -41.83 degrees
i must've done something wrong with the denominator in terms or i and j. i am supposed to get 7.66 degrees. what do i need to fix?
as for part b, how will the fact the charge moves along the +z axis, affect the components of the given magnetic field B?
- #5
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The magnetic field B = 0.3i - 0.4j. Its magnitude is [(0.3)^2 + (0.4)^2]^1/2 = 0.5 T
Using this find the angle.
Related to Magnetic force, finding angle between velocity and magnetic field
1. What is magnetic force?
Magnetic force is a type of force that is exerted on charged particles in a magnetic field. It is the result of the interaction between the magnetic field and the moving charged particles.
2. How is the direction of magnetic force determined?
The direction of magnetic force is determined by the right-hand rule. Point your right thumb in the direction of the charged particle's velocity, and your fingers in the direction of the magnetic field. The direction that your palm is facing is the direction of the magnetic force.
3. How do you find the angle between velocity and magnetic field?
The angle between velocity and magnetic field can be found using the trigonometric function, sine. Simply divide the magnitude of the cross product of velocity and magnetic field by the product of their magnitudes, and then take the inverse sine of this value.
4. What is the unit of magnetic force?
The unit of magnetic force is Newtons (N) in the International System of Units (SI).
5. How is magnetic force used in everyday life?
Magnetic force has many practical applications in everyday life, such as in electric motors, generators, and MRI machines. It is also used in magnetic levitation trains, speakers, and computer hard drives.
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