A boy is pulling a cart by a force of 100N. The frictional force experienced by the cart is 20N. The force causing the motion of
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Answer:
a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m
b) ΔU = -0.000747871 J
c) w = 47.97 rad / s
Explanation:
Given:-
- The area of the circular ring, A = 4.45 cm^2
- The current carried by circular ring, I = 13.5 Amps
- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)
- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)
- The magnetic moment final orientation, vec ( μf ) = -μ k^
- The inertia of ring, T = 6.50×10^−7 kg⋅m2
Solution:-
- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:
μ = N*I*A
Where,
N: The number of turns
I : Current in coil
A: the cross sectional area of coil
- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):
μ = 1*( 13.5 ) * ( 4.45 / 100^2 )
μ = 0.0060075 A-m^2
- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:
vec ( τi ) = vec ( μi ) x vec ( B )
= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )
= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
- Perform cross product:
- The initial torque ( τi ) is written as follows:
vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )
- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):
- The initial potential energy stored in the circular ring ( Ui ) is:
Ui = - vec ( μi ) . vec ( B )
Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]
Ui = - [(-0.000605556 + 0.00011)]
Ui = 0.000495556 J
- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :
Uf = - vec ( μf ) . vec ( B )
Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]
Uf = -0.000252315 J
- The decrease in magnetic potential energy of the ring is arithmetically determined:
ΔU = Uf - Ui
ΔU = -0.000252315 - 0.000495556
ΔU = -0.000747871 J
Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.
- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.
- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:
Ui + Eki = Uf + Ekf
Where,
Eki : The initial kinetic energy ( initially at rest ) = 0
Ekf : The final kinetic energy at second position
- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.
-ΔU = Ekf
0.5*T*w^2 = -ΔU
w^2 = -ΔU*2 / T
Where,
w: The angular speed at second position
w = √(0.000747871*2 / 6.50×10^−7)
w = 47.97 rad / s
Answer: The velocity at different marked time points are given as
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
Explanation:
The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal
axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.
When the tangent of the line is parallel to the horizontal axis, the velocity is 0.
From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
QED!
