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finlep [7]
3 years ago
15

A surveyor is using a magnetic compass 6.4 m below a power line in which there is a steady current of 120 A. (a) What is the mag

netic field in microteslas at the site of the compass due to the power line? (b) Will this field interfere seriously with the compass reading? The horizontal component of Earth’s magnetic field at the site is 20 µT.
Physics
2 answers:
Sergio [31]3 years ago
7 0

Answer:

Explanation:

distance between compass and the wire, r = 6.4 m

current in the wire, i = 120 A

(a)

The magnetic field due to the current carrying conductor is

B = \frac{\mu_{0}}{4\pi}\times \frac{2i}{r}

B = 10^{-7}\times \frac{2\times 120}{6.4}

B = 3.75 x 10^-6 Tesla

(b)

The magnetic field due to the earth is

B' = 20 micro Tesla = 20 x 10^-6 tesla

Yes, the magnetic filed due to current carrying wire interfere the magnetic compass.

Mademuasel [1]3 years ago
6 0

Answer:

Explanation:

Magnetic field due to a long current carrying conductor

μ₀ / 4π x 2i / r  ( i = current , r = distance of point from wire )

= 10⁻⁷ x 2 x 120 / 6.4  ( i = 120 A , r = 6.4m )

= 37.5 x 10⁻⁷ T .

= 3. 75 X 10⁻⁶ T .

= 3.75 µT.

b )

The direction of this field will be horizontal hence it will affect magnetic needle.

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kondor19780726 [428]

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7 0
3 years ago
Two blocks of masses 3.0 kg and 5.0 kg are connected by a spring and rest on a frictionless surface. They are given velocities t
miskamm [114]

Answer:

-0.7 m/sec

Explanation:

Mass of first block = m1 =3.0 kg

Mass of second block = m2= 5.0 kg

Velocity of first block = V1= 1.2 m/s

Velocity of second block = V2 = ?

Momentum of Center of mass MVcom  is sum of both blocks momentum and is given by

MVcom= m1v1+m2v2

Where

M= mass of center of mass

Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)

Putting values, we get;

0= 3×1.2+5v2

==> v2=  3.6/5= - 0.7 m/s

-ve sign indicates that block 2 is moving in opposite direction of block 1

3 0
2 years ago
A rocket is fired at a 45° angle, what is the direction of the horizontal velocity vector at the peak height?
Bess [88]

Answer:

B: Horizontally to the left

Explanation:

Horizontal velocity is always constant throughout the entire trajectory of the rocket and acts in the horizontal direction in which the rocket was launched. This is because gravity only acts in the downwards vertical direction.

So in order words at peak height, horizontal velocity is in the horizontal direction in which the rocket was launched.

So if it was to the left, then direction is left but if right, then direction is right.

Looking at the options, the most appropriate will be:

Horizontally to the left

7 0
2 years ago
A 4.4 nC charge exerts a repulsive force of 36 mN on a second charge which is located
zhenek [66]

The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.

They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.

The given data in the problem is;

q₁  is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C

F is the repulsive force = 36 mN =36 ×10⁶ N

d is the distance = 0.70 m

The Coulomb force is found as;

\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241  \times 10^{-19 } \ C

Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.

To learn more about Coulomb's law, refer to the link;

brainly.com/question/1616890

#SPJ2

6 0
2 years ago
Two identical point charges are 3.00 cm apart. find the charge on each of them if the force or repulsion is 4.00 x 10^-7. (Use C
DanielleElmas [232]

Answer:

Charge on each is 2 x 10⁻¹⁰.

Explanation:

We know that Force between two point charges is given b the Coulomb's law as:

F = kq₁q₂/r^2

k = 9 x 10^9

r = 3.00 cm

= 0.03 m

q₁ = q₂

F = 4.00 x 10^-7

Rearranging the formula, we get:

F = k q²/r²

q² = Fr²/k

q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)

q² = 4 x 10⁻²⁰

q = 2 x 10⁻¹⁰

As there is force of repulsion between the charges, the charges must be both positive or both negative.

3 0
3 years ago
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