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tatiyna
3 years ago
9

A current carrying circular loop of wire lies flat on a table top. When viewed from above, the current moves around the loop in

a counterclockwise sense. What is the direction of the magnetic field caused by this current, inside the loop? The magnetic field A) circles the loop in a counterclockwise direction. B) points straight up. C) points straight down. D) circles the loop in a clockwise direction. E) points toward the east.
Physics
1 answer:
Yanka [14]3 years ago
7 0

Answer:

Option (b)

Explanation:

Maxwell right hand thumb rule: if we hold a current carrying conductor in our right hand such taht the thumb indicates the direction of current then the curling of fingers give the direction of magnetic field around the conductor and vice versa.

Here the current is counterclockwise so by use of Maxwell right hand thumb rule the direction of magnetic field is straight up.

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Explain what happens inside the radiation zone that causes the photon to take so long to exit the sun
DaniilM [7]
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6 0
3 years ago
how many kilograms off mercury would fill a 5litre container if the density of mercury is 13.6grams per cm3
Sidana [21]

Answer:

68kg

Explanation:

1 cm^3 is the same as 1 mL and there are 5000mL in 5L

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7 0
3 years ago
A wire perpendicular to the screen carries a current in the direction shown.
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A) Up is the direction of the magnetic field at point Z.

4 0
3 years ago
Read 2 more answers
What orientation of velocity and acceleration will cause something to initially speed up?
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-- If acceleration and velocity are in the same direction,
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7 0
3 years ago
Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm
Ghella [55]

Answer:

Total electric potential, V=1.32\times 10^6\ volts

Explanation:

It is given that,

First charge, q_1=3\ \mu C=3\times 10^{-6}\ C

Second charge, q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C

Distance of first charge from origin, r_1=1.15\ cm=0.0115\ m

Distance of second charge from origin, r_2=2\ cm=0.02\ m

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}

V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})

V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})

V = 1321826.08 V

or

V=1.32\times 10^6\ volts

So, the total electric potential at the origin is 1.32\times 10^6\ volts. Hence, this is the required solution.

3 0
3 years ago
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