An infinitely long wire carries a current I=100amp . Below the wire a rod of length L=10cm is forced to move at a constant speed v=5m/s along horizontal conducting rails. The rod and rails form a conducting loop. The rod has resistance of R=0.4 ohms . The rails have neglibible resistance. The rod and rails are a distance a=10mm from the wire and in its non-uniform magnetic field as shown. What is the magnitude of the emf induced in the loop in volts?
Let x be the distance from the right end of the rails to the rod.
Magnetic field of the long straight wire is
B=μ₀I/2πr
If the infinitesimal horizontal strip of length x and width dr, parallel to the wire and
a distance r from it => area A = x dr and the flux is
dΦ=BdA= (μ₀I/2πr)xdr
Φ=∫ dΦ= (μ₀Ix/2π)∫dr/r (limits: from ‘a’ to ‘a+L’) = (μ₀Ix/2π) ln{(a+L)/a}
ℰ=dΦ/dt = (μ₀I/2π) (dx/dt) ln{(a+L)/a} =(μ₀Iv/2π) ln{(a+L)/a}=
=(4π•10⁻⁷•100•5/2π) ln{(1+10)/1}=10⁻⁴•ln11=2.4•10⁻⁴ V
To find the magnitude of the emf induced in the loop, we can use Faraday's Law of electromagnetic induction.
1. First, we need to determine the magnetic field at the position of the conducting loop. Since the wire carrying the current is infinitely long, we can use the Biot-Savart Law to calculate the magnetic field at a distance a from the wire.
The formula for the magnetic field due to an infinitely long wire is given by:
B = (μ₀ * I) / (2π * r)
Where B is the magnetic field, μ₀ is the permeability of free space, I is the current in the wire, and r is the distance from the wire.
Plugging in the given values, we get:
B = (4π * 10^-7 T.m/A * 100 A) / (2π * 0.01 m)
= (4 * 10^-6 T.m/A) / (0.02 m)
= 0.2 T
2. Now we can calculate the emf induced in the loop using Faraday's Law of electromagnetic induction, which states that the emf induced in a loop is equal to the rate of change of magnetic flux through the loop.
The formula for the emf induced is given by:
emf = -dΦ/dt
Where emf is the electromotive force (or voltage), dΦ is the change in magnetic flux, and dt is the change in time.
In this case, since the loop is moving at a constant speed v, the change in time dt is equal to the length L divided by the speed v.
So, dt = L / v = 0.1 m / 5 m/s = 0.02 s
3. Next, we need to determine the magnetic flux through the loop. The magnetic flux is given by the product of the magnetic field and the area enclosed by the loop.
In this case, the area enclosed by the loop is equal to the length L multiplied by the distance a.
So, Φ = B * A = 0.2 T * 0.1 m * 0.01 m = 2 * 10^-4 T.m²
4. Finally, we can calculate the emf induced in the loop by taking the negative derivative of the magnetic flux with respect to time.
emf = -dΦ/dt = -(2 * 10^-4 T.m² / 0.02 s)
= - 1 * 10^-2 V
Therefore, the magnitude of the emf induced in the loop is 1 * 10^-2 volts or 0.01 volts.