Given Information:
Length of wire = 132 cm = 1.32 m
Magnetic field = B = 1 T
Current = 2.2 A
Required Information:
(a) Torque = τ = ?
(b) Number of turns = N = ?
Answer:
(a) Torque = 0.305 N.m
(b) Number of turns = 1
Explanation:
(a) The current carrying circular loop of wire will experience a torque given by
τ = NIABsin(θ) eq. 1
Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.
We know that area of circular loop is given by
A = πr²
where radius can be written as
r = L/2πN
So the area becomes
A = π(L/2πN)²
A = πL²/4π²N²
A = L²/4πN²
Substitute A into eq. 1
τ = NI(L²/4πN²)Bsin(θ)
τ = IL²Bsin(θ)/4πN
The maximum toque occurs when θ is 90°
τ = IL²Bsin(90)/4πN
τ = IL²B/4πN
torque will be maximum for N = 1
τ = (2.2*1.32²*1)/4π*1
τ = 0.305 N.m
(b) The required number of turns for maximum torque is
N = IL²B/4πτ
N = 2.2*1.32²*1)/4π*0.305
N = 1 turn
Do not worry if you don't recognize both parts of the problem at this point. If you recognize the dynamics problem,<span> On the other hand, if you recognize this as a kinematics problem you will quickly see that you need to find angular acceleration before you can begin and so will need to do that pre-step first.</span>
m = mass of the person = 82 kg
g = acceleration due to gravity acting on the person = 9.8 m/s²
F = normal force by the surface on the person
f = kinetic frictional force acting on the person by the surface
μ = Coefficient of kinetic friction = 0.45
The normal force by the surface in upward direction balances the weight of the person in down direction , hence
F = mg eq-1
kinetic frictional force on the person acting is given as
f = μ F
using eq-1
f = μ mg
inserting the values
f = (0.45) (82) (9.8)
f = 361.6 N
Answer:
letter B
none zero digit are significant figures
