An ideal toroidal solenoid (Figure 1) has inner radius r1 = 15.1 cm and outer radius r2 = 18.3 cm. The solenoid has 180 turns an
d carries a current of 8.40 A.
a) What is the magnitude of the magnetic field at 12.6 cm from the center of the torus?
b) What is the magnitude of the magnetic field at 16.3 cm from the center of the torus?
c) What is the magnitude of the magnetic field at 20.7 cm from the center of the torus?
a) What is the magnitude of the magnetic field at 12.6 cm from the center of the torus?
b) What is the magnitude of the magnetic field at 16.3 cm from the center of the torus?
c) What is the magnitude of the magnetic field at 20.7 cm from the center of the torus?
2 answers:
Answer:
Part A & C: 0 T
Part B: 1.855*10^-3 T
Explanation:
The formula that models the magnetic field of a toroid is B=μ0*N*I/2π*r .
Note: Toroids keep their magnetic field rotating within the coils that carry current.
Part A & C: Thus the B field magnitude 12.6 cm & 20.7 cm away from the center is 0 T.
Part B:
B=1.855*10^-3 T
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Answer:
V = 49.05 [m/s]
Explanation:
We can easily find the result using kinematics equations, first, we will find the distance traveled during the 5 seconds.
where:
Yo = initial position = 0
y = final position [m]
Vo = initial velocity = 0
t = time = 5 [s]
g = gravity aceleration = 9.81 [m/s^2]
The initial speed is zero, as the body drops without imparting an initial speed. Therefore:
y = 0 + (0*5) + (0.5*9.81*5^2)
y = 122.625[m]
Now using the following equation we can find the speed it reaches during the 5 seconds.
Answer:
A. α = 94.4 rad/s
B. a = 28.32 m/s
C. N = 34N
D. α = 94.4 rad/s
a = 28.32 m/s
N = 44.4 N
Explanation:
part A:
using:
∑T = Iα
where T is the torque, I is the moment of inertia and α is the angular momentum.
firt we will find the moment of inertia I as:
I =
Where M is the mass and R is the radius of the wheel, then:
I =
I = 0.36 kg*m^2
Replacing on the initial equation and solving for α, we get::
∑T = Iα
Fr = Iα
34 N = 0.36α
α = 94.4 rad/s
part B
we need to use this equation :
a = αr
where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:
a = (94.4)(0.3 m)
a = 28.32 m/s
part C
Using the laws of newton, we know that:
N = T
where N is the force that the axle exerts on the wheel part and T is the tension of the cord
so:
N = 34N
part D
The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:
-Mg + N +T = 0
Then, solving for N, we get:
N = -T+Mg
N = -34 + (8 kg)(9.8)
N = 44.4 N