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2 years ago

Calculate the electric field strength at a point at which a test charge of 0.30 coulombs experiences a force of 5.0 newtons.

Physics

1 answer:

SVEN [57.7K]2 years ago

5 0

The strength of electric field E is 17 N / C.

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<u>Explanation:</u>

Electric field strength is defined as the force per unit charge acting at a point in the given field. The equation for the strength of the electric field is given by

E = F / q

where E represents the electric field strength,

F represents the force in newton,

q represents the charge in coulomb.

Given the charge q = 0.30 coulombs

force F = 5.0 N

Electric field strength E = force / charge

= 5.0 / 0.30

E = 16.66 = 17 N / C.

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2 years ago

A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.

zhannawk [14.2K]

Answer:

(a) 1.58 V

(b) 0.0126 Wb

Solution:

As per the question:

No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = $7.50\times 10^{- 3}\ H$

Current in the coil, i = $1680cos[\frac{\pi t}{0.0250}]$ A

where

$i_{max} = 1680\ mA = 1.680\ A$

Now,

(a) To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

$e = \frac{Ldi}{dt}$

$e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]$

$e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]$

For maximum emf, $sin\theta$ should be maximum, i.e., 1

Now, the magnitude of the maximum emf is given by:

$|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V$

(b) To calculate the maximum average flux,we know that:

$\phi_{m, avg} = L\times i_{max} = 7.50\times 10^{- 3}\times 1.680 = 0.0126\ Wb$

$e = e_{o}sin{\pi t}{0.0250}$

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3 years ago

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Helen [10]

Answer:

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2 years ago