A circular conducting ring with radius r0=0.0420 m lies in the x y -plane in a region of uniform

A circular conducting ring with radius r0=0.0420 m lies in...

A circular conducting ring with radius $r_{0}=0.0420 \mathrm{m}$ lies in the $x y$ -plane in a region of uniform magnetic field $\vec{B}=B_{0}\left[1-3\left(t / t_{0}\right)^{2}+\right.$ 2$\left(t / t_{0}\right)^{3} \hat{\boldsymbol{k}} .$ In this expression, $t_{0}=$ 0.0100 $\mathrm{s}$ and is constant, $t$ is time, $\hat{\boldsymbol{k}}$ is the unit vector in the $+z$ - direction, and $B_{0}=0.0800 \mathrm{T}$ and is constant. At points $a$ and $b($ Fig. $\mathrm{P} 29.64)$ there is a small gap in the ring with wires leading to an external circuit of resistance $R=12.0 \Omega .$ There is no magnetic field at the location of the external circuit. (a) Derive an expression, as a function of time, for the total magnetic flux $\Phi_{B}$ through the ring. (b) Determine the emf induced in the ring at time $t=5.00 \times 10^{-3}$ s. What is the polarity of the emf? (c) Because of the internal resistance of the ring, the current through $R$ at the time given in part (b) is only 3.00 mA. Determine the internal resistance of the ring. (d) Determine the emf in the ring at a time $t=1.21 \times 10^{-2}$ s. What is the polarity of the emf? (e) Determine the time at which the current through $R$ reverses its direction.

A circular conducting ring with radius $r_{0}=0.0420 \mathrm{m}$ lies in the $x y$ -plane in a region of uniform magnetic field $\vec{B}=B_{0}\left[1-3\left(t / t_{0}\right)^{2}+\right.$ 2$\left(t / t_{0}\right)^{3} \hat{\boldsymbol{k}} .$ In this expression, $t_{0}=$ 0.0100 $\mathrm{s}$ and is constant, $t$ is time, $\hat{\boldsymbol{k}}$ is the unit vector in the $+z$ - direction, and
$B_{0}=0.0800 \mathrm{T}$ and is constant. At points $a$ and $b($ Fig. $\mathrm{P} 29.64)$ there is a small gap in the ring with wires leading to an external circuit of resistance $R=12.0 \Omega .$ There is no magnetic field at the location of the external circuit. (a) Derive an expression, as a function of time, for the total magnetic flux $\Phi_{B}$ through the ring. (b) Determine the emf
induced in the ring at time $t=5.00 \times 10^{-3}$ s. What is the polarity of the emf? (c) Because of the internal resistance of the ring, the current through $R$ at the time given in part (b) is only 3.00 mA. Determine the internal resistance of the ring. (d) Determine the emf in the ring at a time $t=1.21 \times 10^{-2}$ s. What is the polarity of the emf? (e) Determine the time at which the current through $R$ reverses its direction.

Key Concepts

Lenz's Law

Lenz's Law provides the direction (polarity) of the induced emf and current by stating that the induced current will flow in such a way as to oppose the change in magnetic flux that produced it. This concept is important for determining how the induced currents behave in response to changes in the applied magnetic field, allowing predictions of current direction and voltage polarity in the circuit.

Ohm's Law and Circuit Analysis

Ohm's Law relates the current running through a circuit to the voltage (emf) across it and the total resistance present. In the context of the problem, the circuit features both an external resistor and an internal resistance within the conducting ring, and understanding how these resistances divide the induced emf is necessary to analyze the observed current values and compute the internal resistance.

Magnetic Flux

Magnetic flux is the measure of the magnetic field passing through a given area, which in the case of a loop, is determined by the product of the magnetic field strength perpendicular to the loop and the area of the loop. This concept is crucial as it quantifies the amount of magnetic influence present through the loop, and any change in this flux over time is the source of induced electromotive force (emf).

Faraday's Law of Electromagnetic Induction

Faraday's Law states that a time-varying magnetic flux through a closed loop induces an emf in the loop, with the induced emf being equal to the negative of the rate of change of magnetic flux. This principle is central to the problem since the temporal variation of the applied uniform magnetic field leads to a changing flux through the circular ring and therefore to the generation of an emf.

Transcript

00:02 Hi, in the given problem here this is a circular loop of a conducting wire which is lying in the x, y plane and there is a magnetic field in the direction of positive z -axis.

00:30 The given parameters are for the radius of this loop are not this is equal to 0 .0 420 meter and this magnetic field which is varying with the time in the direction of positive x x x that is given as b bar is equal to b not 1 minus 3 t by t not to the whole square plus two times of t by t not the cube tesla k cap where k is the vector along x xxies the values of these constants means t not is a constant whose value is 0 .0 102 second and b not is having a value of 0 .0 of 0 .0 8800 tesla.

01:43 The resistance of the circuit which is attached with this coil is given as r is equal to 12 .0.

01:59 In the first part of the problem we have to find an expression for the magnetic flux linked with the coil and we know this is given as the dot product of magnetic field with the area vector as coil is lying in xy plane so we can say this area vector will be given by pi times of our not square k -cap means this area vector is also line in positive xxas in the direction of positive x axis so the expression for this flux will become b a cost 0 degree as magnetic field and area both are lying along positive x x x x so the angle between them is 0 degree so it is simply equal to the product of magnetic field expression with the area so it comes out to be b0 1 minus 3 times of t by t0 to the whole square plus 2 times of t by t0 to the whole cube multiplied by area which is pi r not is square so finally the expression for magnetic flux as a function of time comes out to be b not by r not squared multiplied by one minus three times of t by t by t not square plus two times of t by t not q verb which is as a unit of magnetic flux now in the second part of the problem we have to find the value of emf induced so using faraday's law of electromagnetic induction the expression for emf induced in the coil as a result of rate of change of magnetic flux is given by the time rate of change of magnetic flux the negative of the negative sign is used as per lenses law so this is d5 b by dt means time derivative of the function of magnetic flux it will be so we can say it will become minus b not pi r0 as a constant will come out now we have to differentiate the three terms in this bracket first term is one which is a constant so its derivative with respect to time will come out to be zero and then the derivative of three times of t t not will be constant and then it is derivative derivative of two times of tq this is three times of t square and this is two times of tq t not will be constant so these derivatives will become 0 minus 3 times of t not square then the derivative of t square is 2 times of t plus 2 divided by t not cube and then the derivative of t cube which will become 3 times of t square so finally it will come out to be minus b b not pi 0 squared 6 by t0 cube multiplied by t squared minus 6 by t not squared minus 6 by t not is square multiplied by t now we have to find this emf induced at a time of 5 multiplied by 10 bashed power minus 3 second at this much second we have to find the emf induced so putting all the various of the emf and constants, d0, d0, and time, all the things are known.

06:34 So when we put all these values, the emf, induced emf comes out to be, this is minus 0 .08 for magnetic field multiplied by 3 .14, then for radius, this is 0 .04 to the whole square then in the bracket this is six by 0 .01 to the cube multiplied by 5 into 10 dashed par minus 3 square minus 6 by 0 .01 square multiplied by 5 into 10 dashed to the bar minus 3 so we get this is minus 4 .4 3 multiplied by 10 dash to power minus 4 in the bracket it is 150 minus 300 which finally comes out to be 6 .65 multiplied by 10 dash to the power minus 2 volt or we can say this is 0 .0665 so this is the answer for the second part of the problem for the emf induced.

08:05 Now in the third part of the problem we have been given the net current induced in the coil and that has been given as 3 millie ampere.

08:22 It will become 3 multiplied by 10 dash to power minus 3 ampere.

08:29 So using the expression for current, using holmes law.

08:33 This is given as emf divided by the resistance of the circuit plus internal resistance of the coil.

08:41 Let it be smaller.

08:43 So finally this r plus r will be given by emf induced divided by current.

08:50 Here this emf induced is 0 .0665 divided by 3 into 10 dash to bar minus 3 which finally comes out to be approximately 22 .2.

09:06 So this internal resistance of the coil will come out to be 22 .2 minus 12 or this is 10 .2 on.

09:23 In the second part of the problem we have to find the direction also of the induced current because we have to find the polarity also.

09:32 So as when we look at this term, if you look only for the flux, this is the expression, this is the bracket for the flux.

09:44 And if we put the values of constant, it means t0 and time 5 into 10 ratio bar minus three second, definitely this bracket comes out to be negative...

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