A thin charged loop with a charge q>0 (uniformly distributed) and mass m is placed on a ramp which forms and angle α=45∘ with the horizontal. Due to the presence of an external homogeneous magnetic field of magnitude B directed into the page, the loop "takes off" after a time t∗. Determine the time t∗ in seconds if
qB/m=1 s^−1.
Assume that the loop rolls without slipping while in contact with the ramp.
is it 0.75 second? please confirm it
it is a current brilliant challenge problem. you should solve it for yourself!
To determine the time t∗, we need to analyze the forces acting on the charged loop and use the concept of torque and rotational motion.
1. Start by analyzing the forces acting on the charged loop:
- Gravitational force (mg): Acts vertically downward.
- Normal force (N): Acts perpendicular to the ramp surface.
- Magnetic force (Fm): Acts perpendicular to the magnetic field and the velocity of the loop.
- Frictional force (f): Acts parallel to the ramp surface.
2. Since the loop rolls without slipping, the frictional force f can be related to the normal force N using the coefficient of friction (μ) and the gravitational force:
f = μN = μmg
3. The magnetic force Fm on a charged loop moving in a magnetic field is given by the equation:
Fm = qvB, where q is the charge, v is the velocity, and B is the magnetic field.
4. The torque (τ) acting on the loop is given by the product of the magnetic force (Fm) and the radius (r) of the loop:
τ = Fm * r = qvB * r
5. Since the loop is rolling without slipping, the linear velocity of the loop (v) can be related to its angular velocity (ω) using the equation:
v = ωr, where r is the radius of the loop.
6. The torque τ can also be related to the moment of inertia (I) and the angular acceleration (α) of the loop:
τ = I * α
7. The moment of inertia (I) for a loop with mass m and radius r is given by the equation:
I = (1/2) * m * r^2
8. The angular acceleration α of the loop can be related to the linear acceleration a of the loop and the radius r using the equation:
α = a / r
9. The linear acceleration a of the loop can be related to the net force on the loop (ΣFnet) using Newton's second law:
ΣFnet = ma, where ΣFnet = f - mg = μmg - mg
10. Equating the torque equations from step 4 and step 6, we get:
qvB * r = I * α
11. Substituting the equations from step 5, step 8, and step 9 into the torque equation in step 10, we can obtain an expression for the angular acceleration α:
qvB * r = (1/2) * m * r^2 * (a / r)
qvB = (1/2) * m * a
12. Rearranging the equation in step 11, we can solve for the linear acceleration a:
a = (2 * qvB) / m
13. We know that acceleration (a) is the rate of change of velocity (dv) with respect to time (dt). Since the velocity of the loop starts from 0 and increases over time, we can integrate the acceleration equation alongside the appropriate limits to obtain the time t∗:
∫(0 to v∗) dv / ∫(0 to t∗) dt = a
∫(0 to v∗) dv = a * ∫(0 to t∗) dt
14. Integrate the left side of the equation:
[v] from 0 to v∗ = a * (t∗ - 0)
v∗ - 0 = a * t∗
15. Substitute the expression for acceleration a from step 12 into the equation in step 14:
v∗ = (2 * qvB / m) * t∗
16. Since the loop takes off when its velocity v∗ reaches a value equal to the circumference of the loop, we can equate it to 2πr:
v∗ = 2πr
17. Substitute the expression for velocity v∗ from step 16 into the equation in step 15:
(2 * qvB / m) * t∗ = 2πr
18. Rearrange the equation in step 17 to solve for the time t∗:
t∗ = (m * r * 2π) / (2 * qB)
19. Substitute the given values of qB/m = 1 s^−1 and α=45∘ into the equation in step 18:
t∗ = (m * r * 2π) / (2 * (1 s^−1) * B)
Now, you can plug in the actual values of mass (m), radius (r), and the magnetic field (B) to calculate the required time t∗ in seconds.