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kow [346]
3 years ago
7

A conductor is formed into a loop that encloses an area of 1.0 m2. The loop is orientedat a 30.0° angle with the xy-plane. A var

ying magnetic field is oriented parallel to thez-axis. If the maximum emf induced in the loop is 25.0 V, what is the maximum rate atwhich the magnetic field strength is changing
Physics
1 answer:
ivolga24 [154]3 years ago
3 0

Answer:

29 T/s

Explanation:

Given:

Area enclosed by loop, A = 1 m²

Maximum induced emf, e = 25.0 V

loop is oriented at an angle 30° with the x-y plane.

varying magnetic field is oriented parallel to z-axis.

we have to find \frac{dB}{dt}_{max}

Induced emf is given by:

e =-A.cos\theta .\frac{dB}{dt}\\ \Rightarrow \frac{dB}{dt}=-\frac{e}{Acos\theta} = -\frac{25}{1\times cos30^o} = -28.86T/s≈29T/s

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3 years ago
Pls help asap. will mark as brainliest if correct
Deffense [45]

Answer:

Hey mate, here is your answer answer. Hope it helps you

Explanation:

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A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

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Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

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Now,

Electric field at the mid-point due to charge Q' is given by:

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\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

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3 years ago
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