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3 years ago

describe how you can use a compass needle to determine the direction of a magnetic field at a given location

1 answer:

8 0

The compass doesn’t give you the value of the net magnetic field, just the direction. So, how do you get the magnitude of a particular field from this? The trick is to assume the value of the Earth’s magnetic field and the direction of the compass. Let’s assume that at this location on the Earth, the magnetic field is pointing directly North with a horizontal component of about 2 x 10-5 T.

Now suppose that I do something to create a magnetic field in a known direction and perpendicular to the horizontal component of the Earth’s magnetic field. Here is an example where I put a current carrying wire right over the compass needle. Since the compass is underneath the wire, the magnetic field due to the wire will be 90° to the Earth’s magnetic field.

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The earth’s radius is 6.37 × 10 6 m ; it rotates once every 24 hours. What is the earth’s angular speed? Viewed from a point abo

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Answer:

a) the angular speed of the Earth's rotation is 7.272 × 10⁻⁵ rad/s.

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the angular velocity is positive.

c) Earth's speed at a point on the equator is 463.23 m/s.

d) the speed of a point on the earth’s surface halfway between the equator and the pole is 231.61 m/s.

Explanation:

a) The angular speed of the Earth's rotation is:

ω = 2π / T

where

T is the period

ω = 2π / (24 hr × (3600 s / 1 hr))

ω = 7.272 × 10⁻⁵ rad/s

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the angular velocity is positive.

c) Earth's speed at a point on the equator is:

v = r × ω

v = (6.37 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

v = 463.23 m/s

d) The radius of the circle in which the point moves is half of Earth's radius. Therefore,

r = 1/2(6.37 × 10⁶ m)

r = 3.19 × 10⁶ m

Therefore, the speed of a point on the earth’s surface halfway between the equator and the pole is:

v = r × ω

v = (3.19 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

v = 231.61 m/s

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3 years ago

You have a remote-controlled car that has been programmed to have velocity v⃗ =(−3ti^+2t2j^)m/s, where t is in s. At t = 0 s, th

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Answer:

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The x-component of the acceleration vector is -3, and the y-component is 40.

Explanation:

The displacement vector of the car with velocity

$\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s$