In the above case we can say that power given by external agent to pull the rod must be equal to the power dissipated in the form of heat due to magnetic induction.
Part a)
when we pull the rod with constant speed then power required will be product of force and velocity
here we will have
P = 4 W
v = 4 m/s
now we will have
So external force required will be 1 N
PART B)
now in order to find magnetic field strength we can say
here we know that induced EMF in the wire is E = vBL
so power due to induced magnetic field is given by
by solving above equation we will have
The unit of electric current is the 'ampere'.
That's the current in the circuit when one coulomb of charge
flows past any fixed point every second.
Answer:
x(t) = d*cos ( wt )
w = √(k/m)
Explanation:
Given:-
- The mass of block = m
- The spring constant = k
- The initial displacement = xi = d
Find:-
- The expression for displacement (x) as function of time (t).
Solution:-
- Consider the block as system which is initially displaced with amount (x = d) to left and then released from rest over a frictionless surface and undergoes SHM. There is only one force acting on the block i.e restoring force of the spring F = -kx in opposite direction to the motion.
- We apply the Newton's equation of motion in horizontal direction.
F = ma
-kx = ma
-kx = mx''
mx'' + kx = 0
- Solve the Auxiliary equation for the ODE above:
ms^2 + k = 0
s^2 + (k/m) = 0
s = +/- √(k/m) i = +/- w i
- The complementary solution for complex roots is:
x(t) = [ A*cos ( wt ) + B*sin ( wt ) ]
- The given initial conditions are:
x(0) = d
d = [ A*cos ( 0 ) + B*sin ( 0 ) ]
d = A
x'(0) = 0
x'(t) = -Aw*sin (wt) + Bw*cos(wt)
0 = -Aw*sin (0) + Bw*cos(0)
B = 0
- The required displacement-time relationship for SHM:
x(t) = d*cos ( wt )
w = √(k/m)
Answer:
D.)
Explanation:
the current separates on each branch according to the resistance it experience.
