"A rod with length L , mass m and resistance R slides without friction down parallel conducting rail of
negligible resistance . the rails are connected together at bottom, forming a conducting loop with the rod
the plain of the loop makes angle Z wih horizontal , a uniform vertical magnetic field B exist
througth the region perpendicular to the plane . show that the rod acquire a velocity whose magnetude
v = (mgR sinZ)/square of(BLcosZ)
and
show that the rate at which the internal energy of the rod is increasing is equal to the rate at which
the rod is losing gravitational potential energy
To solve this problem, we can use the principles of electromagnetism and conservation of energy. Let's break it down step by step:
Step 1: Magnetic Force on the Rod
The rod is sliding down the conducting rails, which are perpendicular to the magnetic field. According to the electromagnetic induction principle, when a conducting rod moves in a magnetic field, a force is generated on the rod.
First, let's find the magnetic force acting on the rod. The magnetic force (F) is given by the formula:
F = BILsinθ
where B is the magnetic field, I is the current passing through the rod, L is the length of the rod, and θ is the angle between the velocity of the rod and the magnetic field.
Step 2: Current passing through the rod
Since the rails are connected together at the bottom, a current flows through the conducting loop formed by the rod and the rails. The current (I) can be found using Ohm's Law:
I = V/R
where V is the voltage induced across the rod due to the magnetic force and R is the resistance of the rod.
Step 3: Voltage induced across the rod
The voltage induced across the rod is given by Faraday's Law of Electromagnetic Induction:
V = BLVsinθ
where B is the magnetic field, L is the length of the rod, V is the velocity of the rod, and θ is the angle between the velocity of the rod and the magnetic field.
Step 4: Equating the magnetic force and the product of current and resistance
Now, let's equate the magnetic force on the rod (F) to the product of current (I) and resistance (R):
F = IR
Substituting the values we found earlier, we get:
BILsinθ = (BLVsinθ)(V/R)
Simplifying, we get:
mg = V^2/R
Step 5: Finding the velocity
Using the equation from Step 4, we can rearrange it to solve for the velocity of the rod (V):
V^2 = mgR
Taking the square root of both sides gives us:
V = sqrt(mgR)
Step 6: Finalizing the expression for velocity
The problem states that the velocity of the rod is given by:
v = (mgRsinθ) / (BLLcosθ)
We can simplify this expression by substituting V from Step 5:
v = (mgRsinθ) / (BLcosθ)
This is the final expression for the magnitude of the velocity acquired by the rod.
Step 7: Internal Energy and Gravitational Potential Energy
The rate at which the internal energy of the rod is increasing is equal to the rate at which the rod is losing gravitational potential energy.
The gravitational potential energy lost by the rod can be given as:
ΔPE = mgΔh
where m is the mass of the rod, g is the acceleration due to gravity, and Δh is the change in height.
If the rod slides down a height of h, then the change in height is given by:
Δh = hsinθ
Hence, ΔPE = mghsinθ
The rate at which the rod loses gravitational potential energy is given by:
dPE/dt = mghsinθv/t
where v is the velocity of the rod and t is the time taken to slide down the height h.
Substituting the value of v from Step 6, we get:
dPE/dt = mghsinθ[(mgRsinθ) / (BLLcosθ)] / t
Simplifying, we get:
dPE/dt = (mg^2R) / (BLt)
This expression shows that the rate at which the internal energy of the rod is increasing is equal to the rate at which the rod is losing gravitational potential energy.
So, we have derived the expressions for the magnitude of the velocity acquired by the rod and the rate at which the internal energy of the rod is increasing.