What is the emf induced in the loop?

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    Emf Induced Loop
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The discussion focuses on calculating the induced electromotive force (emf) in a loop placed inside a long solenoid. The relevant formula for induced emf is N(deltaBA/deltat), where N is the number of turns in the loop, deltaB represents the change in magnetic field, and deltat is the change in time. The magnetic field inside the solenoid is determined by B = mu0*n*I, with mu0 being the permeability of free space and n the turns per unit length. To find deltaB, the current's time-dependent behavior is factored in, leading to a final expression for the induced emf based on the specified parameters. The induced emf is thus influenced by the loop's turns, the magnetic field changes, and the time interval.
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A long solenoid has n = 400 turns/m and carries a current I = 30 A(1 - e-1.6t/s). Inside the solenoid and coaxial with it is a loop that has a radius R = 0.06 m and consists of N = 250 turns of wire (Fig. 3). What is the emf induced in the loop?

I know that the inuced emf is N(deltaBA/deltat) but I'm not sure what to use when there is an inside coil as well. Or maybe my the equation is wrong all together...please help
 
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BillytheKid said:
A long solenoid has n = 400 turns/m and carries a current I = 30 A(1 - e-1.6t/s). Inside the solenoid and coaxial with it is a loop that has a radius R = 0.06 m and consists of N = 250 turns of wire (Fig. 3). What is the emf induced in the loop?

I know that the inuced emf is N(deltaBA/deltat) but I'm not sure what to use when there is an inside coil as well. Or maybe my the equation is wrong all together...please help
Long solenoids create a uniform magnetic field in their interiors, correct? Well, the equation given for the amperage... this will cause the magnetic field to change with time, your Delta B, which yields an induced EMF in the interior coil.
 


The equation you have is correct. To calculate the induced emf in the loop, you will need to use the formula N(deltaBA/deltat), where N is the number of turns in the loop, deltaB is the change in magnetic field, and deltat is the change in time.

In this case, the magnetic field inside the solenoid is given by B = mu0*n*I, where mu0 is the permeability of free space, n is the number of turns per unit length, and I is the current. Using this, we can calculate the change in magnetic field as deltaB = mu0*n*(I(t2) - I(t1)), where t2 and t1 are the final and initial times respectively.

Substituting the values given in the problem, we get deltaB = (4*pi*10^-7)*(400)*(30*(1-e^-1.6t2/s) - 30*(1-e^-1.6t1/s)).

Now, the change in time can be calculated as deltat = t2 - t1.

Finally, substituting these values in the formula N(deltaBA/deltat), we get the induced emf in the loop as N*[(4*pi*10^-7)*(400)*(30*(1-e^-1.6t2/s) - 30*(1-e^-1.6t1/s))]/(t2-t1).

Therefore, the induced emf in the loop is dependent on the number of turns in the loop, the change in magnetic field inside the solenoid, and the change in time.
 
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