A circular loop of radius 13 cm carries a current of 13 A. A flat coil of radius 0.94 cm, having 58 turns and a current of 1.9 A, is concentric with the loop. The plane of the loop is perpendicular to the plane of the coil. Assume the loop's magnetic field is uniform across the coil. What is the magnitude of (a) the magnetic field produced by the loop at its center and (b) the torque on the coil due to the loop
Answer:
[tex]6.28\times 10^{-5}\ \text{T}[/tex]
[tex]1.92\times 10^{-6}\ \text{Nm}[/tex]
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi 10^{-7}\ \text{H/m}[/tex]
[tex]I_l[/tex] = Current in circular loop = 13 A
[tex]r_l[/tex] = Radius of circular loop = 13 cm
[tex]N[/tex] = Number of turns = 58
[tex]r_c[/tex] = Radius of coil = 0.94 cm
[tex]I_c[/tex] = Current in coil = 1.9 A
[tex]\theta[/tex] = Angle between loop and coil = [tex]90^{\circ}[/tex]
Magnitude of magnetic field in circular loop
[tex]B_l=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B_l=\dfrac{4\pi 10^{-7}\times 13}{2\times 13\times 10^{-2}}\\\Rightarrow B_l=6.28\times 10^{-5}\ \text{T}[/tex]
The magnetic field produced by the loop at its center is [tex]6.28\times 10^{-5}\ \text{T}[/tex].
Torque is given by
[tex]\tau=\pi NI_cr_c^2B_l\sin\theta\\\Rightarrow \tau=\pi 58\times 1.9\times (0.94\times 10^{-2})^2\times 6.28\times 10^{-5}\sin90^{\circ}\\\Rightarrow \tau=1.92\times 10^{-6}\ \text{Nm}[/tex]
The torque on the coil due to the loop [tex]1.92\times 10^{-6}\ \text{Nm}[/tex].
what is the minimum effort required to raise the block?
Answer:
2000 N.
Explanation:
From the question given above, the following data were obtained:
Load (L) = 8000 N
Mechanical advantage (MA) = 4
Effort (E) =?
The mechanical advantage of a machine is simply defined as:
Mechanical advantage = Load / Effort
MA = L / E
With the above formula, we can obtain the effort used to raise the load of 8000 N as follow:
Load (L) = 8000 N
Mechanical advantage (MA) = 4
Effort (E) =?
MA = L / E
4 = 8000 / E
Cross multiply
4 × E = 8000
Divide both side by 4
E = 8000 / 4
E = 2000 N
Thus, the effort used to raised the load is 2000 N.
1 ) when a ball is projected upwords its time of rising is ...............the time of falling .
a) greater than b) smaller than c) equal to d ) double
2 ) when an object falls freely under the effect of gravity , the distance moved is
a ) directly proportional to time
b ) inversely proportional to time
c ) directly proportional to square of time
d ) inversely proportional to square of time.
Answer:
correct answer is C
Explanation:
In this exercise, you are asked to complete the sentences so that the sentence makes sense.
1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down
correct answer C
2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is
y = v₀ t - ½ gt²
v₀ = 0
y = -1/2 g t²
so the position is not proportional to the square of the time
correct answer is C
Calculate the volume occupied by a glass cup having a mass of 260 g knowing that the
density of glass is equal to 2.6 g/cm3
Answer:
100 milliliters
Explanation:
Low air pressure and warm temperatures over land are most likely to result in which of the following weather conditions or events?
A.Clear sunny skies
B. tropical hurricane
C. chance of snow
D.Cloudy skies with rain
Definition of continental polar
cold, dry, and stable air masses .
Find the velocity of an object that has a mass of 500x10-6 kg and a charge of 1pc if it starts from
rest and passes through a potential difference of 120kV
Answer:
v = 4.8 10⁻⁴ m / s
Explanation:
To solve this exercise we can use the concepts of energy. In this case the potential energy is transformed into kinetic energy
U = K
q V = ½ m v²
v = [tex]\sqrt { \frac{2qV}{m} }[/tex]
in the exercise they indicate the value of the charge q₁ = 1 pC = 1 10⁻¹² C
let's calculate
v = [tex]\sqrt{ \frac{2 \ 1 \ 10^{-12} 120 \ 10^3}{500 ^{2} }[/tex]
v = 4.8 10⁻⁴ m / s