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2 years ago

9

Sully uses a battery and a coil of wire to create an electromagnet. Using the same materials, if he wants to increase the streng

th of his electromagnet, he must _____ the number of loops in the coil of wire along its length. If he uses an additional battery, the strength of his electromagnet will _______.
first blank has options: increase, decrease, no change

second blank has the same options

2 answers:

fenix001 [56]2 years ago

6 0

Answer : The correct answer for both the fill in the blank is, increases.

Explanation :

Formula used :

1) $B=\mu_o\times \frac{N\times I}{L}$

B = strength of magnetic filed

N = Number of loops

I = current in the wire

$\mu_o$ = magnetic permeability

$B\propto N$

The more the number of loops in the solenoid more will the strength of three magnetic fields.

2) Also according to Ohm's law:

Voltage is directly proportional to current:

$V\propto I$

With increase in voltage the the current will increase which in return increase the magnetic field in the solenoid.

So, the correct answer for both the fill in the blank is, increases.

Rudiy272 years ago

3 0

Answer is increase for both.

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2 years ago

Read 2 more answers

A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons

shepuryov [24]

Answer:

$v=2.556m/s$

Explanation:

From the conservation of mechanical energy

$K_{E1}+U_1=K_{E2}+U_2$

$\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2$

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Solve to velocity v2

$m*v_2^2=k*x_1^2-k*x_2^2$

$v^2=\frac{k}{m}*(x_1^2-x_2^2)$

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3 years ago

A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

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$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

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