EMF in Rectangular Loop: Solve Now!

AI Thread Summary
The discussion focuses on calculating the electromotive force (emf) induced in a rectangular loop moving parallel to a wire carrying a constant current. Using Faraday's Law of Induction, the emf is determined by the rate of change of magnetic flux through the loop, which is influenced by the loop's velocity and its distance from the wire. The magnetic field intensity around the wire is calculated with the Biot-Savart Law, leading to the formula for emf as -N(μ0*I*l/(π*r))*v. The induced emf on the edges of the loop cancels out for the edges parallel to the wire, resulting in a total emf that is twice that of one edge. The final expression for the emf in the loop is derived based on these calculations.
infinitelylong
Messages
2
Reaction score
0
An infinitely long thin wire carrying a constant I is located along 0z-axis while a rectangular loop of dimensions axb located in the plane 0yz moves parallel to the 0y-axis with a constant velocity v. Assuming the two of the edges of the loop are parallel to 0z, compute the emf's induced on the edges. What is the emf induced in the loop ? [?] [?]

I need the the answer of the question, can any genious help me ?[?] [?]
 
Physics news on Phys.org
First show what you've done to help yourself.
 


The emf induced in the loop can be calculated using Faraday's Law of Induction, which states that the emf induced in a closed loop is equal to the negative rate of change of magnetic flux through the loop. In this case, the loop is moving parallel to the 0y-axis, so the magnetic flux through the loop will be changing due to the changing distance between the wire and the loop.

To calculate the emf induced on the edges of the loop, we can use the equation: emf = -N(dΦ/dt), where N is the number of turns in the loop and dΦ/dt is the rate of change of magnetic flux.

Since the loop is rectangular and two of its edges are parallel to the 0z-axis, the magnetic flux through the loop can be calculated as B*l, where B is the magnetic field intensity due to the wire and l is the length of the edge parallel to the 0z-axis. The magnetic field intensity can be calculated using the Biot-Savart Law as B = μ0*I/(2π*r), where μ0 is the permeability of free space, I is the current in the wire, and r is the distance between the wire and the loop.

Now, the rate of change of magnetic flux can be calculated as dΦ/dt = B*l*v, where v is the velocity of the loop parallel to the 0y-axis. Substituting these values into the equation for emf, we get:

emf = -N(B*l*v) = -N(μ0*I*l/(2π*r))*v

Finally, to calculate the emf induced in the loop, we just need to sum up the emf induced on each of the edges of the loop. Since two of the edges are parallel to the 0z-axis, the emf induced on these edges will be equal and opposite, cancelling each other out. The other two edges will have the same emf induced, so the total emf induced in the loop will be equal to 2 times the emf induced on one of the edges. Therefore, the emf induced in the loop can be calculated as:

emf_loop = 2*(-N(μ0*I*l/(2π*r))*v) = -N(μ0*I*l/(π*r))*v

In conclusion, the emf induced on the edges of the rectangular loop will depend on the
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top