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
Answer:
Spring constant of the spring will be equal to 9.255 N /m
Explanation:
We have given mass m = 0.683 kg
Time taken to complete one oscillation is given T = 1.41 sec
We have to find the spring constant of the spring
From spring mass system time period is equal to 
, here m is mass and K is spring constant
So 
Squaring both side
So spring constant of the spring will be equal to 9.255 N /m
Answer: KE = 62.5J
Explanation:
Given that
Mass of object = 5kg
kinetic energy KE = ?
velocity of object = 5m/s
Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.
i.e K.E = 1/2mv^2
KE = 1/2 x 5kg x (5m/s)^2
KE = 0.5 x 5 x 25
KE = 62.5J
Thus, the object has 62.5 joules of kinetic energy.
Oxygen....................
