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Goshia [24]
3 years ago
5

A single loop of wire with an area of 0.0780 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic

ular to the plane of the loop, and is decreasing at a constant rate of 0.240 T/s .
Requried:
If the loop has a resistance of 0.700Ω. Find the current induced in the loop.
Physics
1 answer:
Katarina [22]3 years ago
4 0

Answer:

The induced current is 26.7 mA

Explanation:

Given;

area of the loop, A = 0.078 m²

initial magnetic field, B₁ = 3.8 T

change in the magnetic field strength, dB/dt = 0.24 T/s

The induced emf is calculated as;

emf = - \frac{d \phi}{dt} \\\\emf = -\frac{dB.A}{dt} \\\\emf = A (\frac{dB}{dt} )\\\\emf = 0.078(0.24)\\\\emf = 0.0187 \ V

The resistance of the loop = 0.7 Ω

The induced current is calculated as;

V = IR\\\\I = \frac{V}{R} = \frac{emf}{R} = \frac{0.0187}{0.7} = 0.0267 \ A = 26.7 \ mA

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Answer:

a

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Explanation:

From the question we are told that

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    The  is distance of the screen from the slit is  D   =  2.60 \ m

    The distance between the central bright fringe and either of the adjacent bright   y  =  1.97 cm  =  1.97 *10^{-2}\ m

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      d sin \tha(\theta ) =  n \lambda

From the question we are told that small-angle approximation is valid here.

So    sin (\theta ) = \theta

=>        d \theta  =  n \lambda

=>        \theta =  \frac{n *  \lambda }{d }

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         y  =  D * sin (\theta )

From the question we are told that small-angle approximation is valid here.

So

       y  =  D * \theta

=>   \theta  =  \frac{ y}{D}

So

     \frac{n *  \lambda }{d } = \frac{y}{D}

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substituting values

       \lambda =  \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }

        \lambda = 1.667 *10^{-6}

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In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     dsin (\theta ) =  n \lambda

    \theta  =  sin^{-1}[\frac{ n  *  \lambda }{d} ]

    \theta  =  sin^{-1}[\frac{ 2  *  1667 *10^{-9}}{ 0.00022} ]

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Answer:

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