The force on a charge moving
with a velocity v in a magnetic
field B is half of the maximum
force when the angle between
wave B is:
A. 0°
B. 90°
C. 45°
D. 30°
The force on a charge moving in a magnetic field is given by the equation F = qvBsinθ, where F is the force, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
To find the angle at which the force is half of the maximum force, we can set F = (1/2)F_max and solve for θ.
(1/2)F_max = qvBsinθ
Since sinθ has a maximum value of 1, we can rewrite the equation as:
(1/2)F_max = qvB
Now we can solve for θ.
(1/2)F_max = qvB
(1/2)(qvB) = qvB
(1/2) = sinθ
sinθ = 1/2
θ = sin^(-1)(1/2)
θ = 30°
Therefore, the angle between the velocity vector and the magnetic field vector when the force is half of the maximum force is 30°.
Answer: D. 30°
The force on a charge moving with a velocity v in a magnetic field B is given by the equation F = qvB sin(theta), where q is the charge of the particle, v is the velocity of the particle, B is the magnetic field, and theta is the angle between the velocity vector and the magnetic field vector.
If we are given that the force on the charge is half of the maximum force, we can set up the following equation:
F = (1/2)F_max
Using the equation F = qvB sin(theta), we can substitute F with (1/2)F_max:
qvB sin(theta) = (1/2)(qvB)
We can now cancel out q and B from both sides:
v sin(theta) = (1/2)v
Next, we can cancel out v from both sides:
sin(theta) = 1/2
To find the value of theta, we can take the inverse sine of both sides:
theta = sin^(-1)(1/2)
Evaluating this expression, we find:
theta = 30°
Therefore, the correct answer is D. 30°.