Question

In: Physics

A single loop of wire with an area of 0.0900m2 is in a uniform magnetic field...

A single loop of wire with an area of 0.0900m2 is in a uniform magnetic field that has an
initial value of 3.8 T, is perpendicular to the plane of the loop, and is decreasing at a
constant rate of 0.190 T/s. 
(a) What emf is induced in this loop? 
(b) If the loop has a resistance of 0.600O, find the current induced in the loop.

Solutions

Expert Solution

Concepts and reason

The concept to be used in this problem is of Faraday’s Law of Electromagnetic Induction.

First, apply Faraday’s Law of Electromagnetic Induction law to find the emf induced in the loop. Then, using the voltage-current relation find the current flowing in the loop.

Fundamentals

The Law of Electromagnetic Induction states that the emf induced in the closed circuit is equal to the change in the magnetic flux through the circuit.

ε=NdϕBdt\varepsilon = - N\frac{{d{\phi _B}}}{{dt}}

Here NN is the number of turns of the coil and ϕB{\phi _B} is the magnetic flux.

The expression of magnetic flux is given by,

ϕB=BA{\phi _B} = BA

Here BB is the magnitude of the magnetic field and AA is the area of the surface.

The voltage-current relation is given by the Ohm's law:

I=VRI = \frac{V}{R}

Here VV is the emf induced in the loop and RR is the resistance of the loop.

(a)

The Law of Electromagnetic Induction states that the emf induced in the closed circuit is equal to the change in the magnetic flux through the circuit.

ε=NdϕBdt\varepsilon = - N\frac{{d{\phi _B}}}{{dt}}

Here NN is the number of turns of the coil and ϕB{\phi _B} is the magnetic flux.

Substitute BABA for ϕm{\phi _m} ,

ε=Nddt(BA)=NAdBdt\begin{array}{c}\\\varepsilon = - N\frac{d}{{dt}}\left( {BA} \right)\\\\ = - NA\frac{{dB}}{{dt}}\\\end{array}

Substitute 1 for NN , 0.09m20.09{\rm{ }}{{\rm{m}}^2} for AA and 0.19Ts10.19{\rm{ T}} \cdot {{\rm{s}}^{ - 1}} for dBdt\frac{{dB}}{{dt}} ,

ε=(1)(0.09m2)(0.19Ts1)=0.0171V=1.71×102V\begin{array}{c}\\\varepsilon = - \left( 1 \right)\left( {0.09{\rm{ }}{{\rm{m}}^2}} \right)\left( {0.19{\rm{ T}} \cdot {{\rm{s}}^{ - 1}}} \right)\\\\ = - 0.0171{\rm{ V}}\\\\{\rm{ = }} - 1.71 \times {10^{ - 2}}{\rm{ V}}\\\end{array}

Hence the magnitude of the induced emf is 1.71×102V1.71 \times {10^{ - 2}}{\rm{ V}} .

(b)

The current flowing in the loop is given by,

I=εRI = \frac{\varepsilon }{R}

Here ε\varepsilon is emf induced in the loop and RR is the resistance of the loop.

Substitute 1.71×102V1.71 \times {10^{ - 2}}{\rm{ V}} for ε\varepsilon and 0.6Ω0.6\Omega for RR ,

I=1.71×102V0.6Ω=2.85×102A\begin{array}{c}\\I = \frac{{1.71 \times {{10}^{ - 2}}{\rm{ V}}}}{{0.6\Omega }}\\\\ = 2.85 \times {10^{ - 2}}{\rm{ A}}\\\end{array}

Ans: Part a

The emf induced in the loop is 1.71×102V1.71 \times {10^{ - 2}}{\rm{ V}} .

Part b

The Current induced in the loop is 2.85×102A2.85 \times {10^{ - 2}}{\rm{ A}} .


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