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Masteriza [31]
3 years ago
12

A constant magnetic field of 0.50 T is applied to a rectangular loop of area 3.0 × 10-3 m2. If the area of this loop changes fro

m its original value to a new value of 1.6 × 10-3 m2 in 1.6 s, what is the emf induced in the loop?
Physics
1 answer:
Nonamiya [84]3 years ago
8 0

Answer:

e= 0.440 mV(millivolt)

Explanation:

Given Magnetic field B= 0.50T

A_{1}=3\times10^{-3}

A_{2}=1.6\times10^{-3}\\

Δt= 1.6sec

induced emf e=\frac{\mathrm{d} \phi }{\mathrm{d} t}

⇒e=B\frac{\mathrm{d} A}{\mathrm{d} t}

e=0.50\times\frac{3\times10^{-3}-1.6\times10^{-3}}{1.6}

e=0.440\times10^{-3} volt

e= 0.440 mV(millivolt)

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Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

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{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

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{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

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