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In Figure, a wire forms a closed circular loop, of radius R = 2.0mand resistance4.0. The circle is centered on a long straight wire; at time t = 0, the current in the long straight wire is 5.0 Arightward. Thereafter, the current changes according toi=5.0A-(2.0As2)t2. (The straight wire is insulated; so there is no electrical contact between it and the wire of the loop.) What is the magnitude of the current induced in the loop at times t > 0?

Short Answer

Expert verified

The induced current in the circle is, i = 0.

Step by step solution

01

Given

  1. The radius of the circular loop R = 2.0 m
  2. The resistance4.0
  3. At time, t = 0, the current in the long straight wire is 5.0 A rightward.
  4. The current changes according toi=5.0A-2.0A/s2t2.
  5. The straight wire is insulated.
02

Determining the concept

Applying the right hand rule, find the net flux produced at any time. Substituting this value in eq.30-4, find the emf induced due to change in the magnetic flux. Finally, applying Ohm鈥檚 law, find the magnitude of the current induced in the loop.

Right Hand Rule states that if we arrange our thumb, forefinger and middle finger of the right-hand perpendicular to each other, then the thumb points towards the direction of the motion of the conductor relative to the magnetic field, the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the induced current.

Faraday's law of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Formulae are as follow:

=-dBdti=R

Where,Bis magnetic flux, i is current, R is resistance, 饾渶 is emf.

03

Determining the induced current in the circle

According to Faraday鈥檚 law, the emf is induced due to change in the magnetic flux is,

=-dBdt.........................(30-4)

Since,the current in the long straight wire is rightward. Therefore, according to the right-hand rule the field due to the current in the straight wire is out of the page in the upper half of the circle and is into the page in the lower half of the circle.

Hence, the net flux produced at any time is must be zero.

That is,

B=0

Substituting in Eq.30-4,

=-d(0)dt=0

According to Ohm鈥檚 law,

i=R

Therefore, the induced current in the circle is,

i=0Ri=0

Hence, the induced current in the circle is, i = 0.

Therefore, by using the concept of Right hand rule, Faraday鈥檚 law and Ohm鈥檚 law, the current induced in the circle can be determined.

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Most popular questions from this chapter

Suppose the emf of the battery in the circuit shown in Figure varies with time t so that the current is given by i(t) = 3.0 +5.0 t , where i is in amperes and t is in seconds. Take R=4.0andL=5.0H, and find an expression for the battery emf as a function of t. (Hint: Apply the loop rule.)

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In Fig. 30-26, a wire loop has been bent so that it has three segments: segment bc(a quarter-circle), ac(a square corner), and ab(straight). Here are three choices for a magnetic field through the loop:

(1)B1=3i^+7j^-5tk^,(2)B2=5ti^-4j^-15k^,(3)B3=2i^-5tj^-12k^,

where Bis in milliteslas and tis in seconds. Without written calculation, rank the choices according to (a) the work done per unit charge in setting up the induced current and (b) that induced current, greatest first. (c) For each choice, what is the direction of the induced current in the figure?

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