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4 years ago

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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.

2 answers:

Sergio [31]4 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]4 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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A solenoidal coil with 21 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 22.0 cm long

Thepotemich [5.8K]

Answer:

(a)The average of magnetic flux trough each turns is $3.63 \times 10^{-8}$ wb.

(b) The magnitude of mutual inductance of the two solenoid is $1.596 \times 10^{-5}$ H.

(c)The magnitude of emf induced in the outer solenoid by changing current in the inner solenoid is 0.027132 v .

Explanation:

Given that,

The number of turns of the solenoid=N₁= 350

The diameter of the solenoid=D= 2.20 cm=0.022

The length of the solenoid is = $l$ = 22.0 cm=0.22 m

The second solenoid with $N_2$ = 21 turns which is wound around the solenoid at its center.

The current in the inner solenoid is $I_1$ = 0.150 A and is increasing at a rate of 1700 A/s i.e $\frac{dI_1}{dt}= 1700$ A/s

(a)

The formula of magnetic field of a solenoid is

$B=\mu_0 nI=\frac{\mu_0 N_1I_1}{l}$

$=\frac{(4\pi \times 10^{-7} T.m/A) (350)(0.15 \ A)}{0.220 \ m}$

$\approx 3.0 \times 10^{-4}$ T

So, the magnetic flux of each turns

$\phi = BA=B \pi (\frac d2)^2$

$=(3.0\times 10^{-4}\ T)\pi (\frac {0.022}{2} \ m)^2$

$=1.14 \times 10^{-7}$ wb

The average of magnetic flux trough each turns is $3.63 \times 10^{-8}$ wb.

(b)

Since the both coil wound tightly. So, the magnitude magnetic flux though each turns of both coils is same.

The mutual inductance of the two solenoid is

$M=\frac{N_2\phi}{I_1}$

$=\frac{(21)(1.14 \times 10^{-7}\ wb)}{0.15\ A}$

$=1.596 \times 10^{-5}$ H

(c)

The formula of emf is

$\varepsilon =-M\frac{dI_1}{dt}$

$=-(1.596\times 10^{-5}\ H) (1700 \ A/s)$

$=27.132 \times 10^{-3}$ v

=0.027132 v

The emf induced in the outer solenoid by changing current in the inner solenoid is 0.027132 v .

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Explanation:

The value of acceleration due to gravity on the surface of earth is given as:

g = GM/R² -------------------- equation 1

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G = Universal Gravitational Constant

M = Mass of Earth

R = Radius of Earth

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gₓ = GMₓ/Rₓ²

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Rₓ = Radius of Alien Planet = 1.9 R

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gₓ = 0.66 GM/R²

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