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ira [324]
3 years ago
12

A compass originally points North; at this location the horizontal component of the Earth's magnetic field has a magnitude of 2e

-5 T. A bar magnet is aligned East-West, pointing at the center of the compass. When the center of the magnet is 0.25 m from the center of the compass, the compass deflects 70 degrees. What is the magnetic dipole moment of the bar magnet?
Physics
1 answer:
astraxan [27]3 years ago
8 0

Answer:

μ =5.40 A-m²

Explanation:

The components of the net magnetic field are the magnetic field of the dipole and the magnetic field of Earth, then from the right triangle, the deflection angle is computed by

tan θ = Bdipole / Bearth     ⇒   Bdipole = Bearth* tan θ  

Bdipole = 2e-5 T*tan 70° = 5.49e-5 T

The magnetic field at the location of the compass due to the dipole has a magnitude

Bdipole = (μ₀/4π)(2μ/r³)    ⇒    μ = Bdipole r³ / 2(μ₀/4π)

μ = (5.49e-5 T)(0.27m)³ / 2(1 × 10−7 T m² /(C m/s)) = 5.40 A-m²

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3 years ago
What three sprites are needed to create the Space Invaders game that was discussed in the unit? spaceship, laser, enemy gnome, a
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3 0
2 years ago
A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

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I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

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4 0
3 years ago
A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.
fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force N= mg cos 22

\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N





8 0
3 years ago
Read 2 more answers
What is the molar mass of a gas if 1.30g of the gas has a volume of 245mL at STP? ...?
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First, we assume this as an ideal gas so we use the equation PV=nRT. Then, we use the conditions at STP that would be 1 atm and 273.15 K. We calculate as follows:

PV= nRT
PV= mRT/MM

1 atm (.245 L) =1.30(0.08206)(273.15) / MM
MM = 118.94 g/mol <--- ANSWER
5 0
3 years ago
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