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KIM [24]
3 years ago
7

An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 1.80 T field with his f

ingers pointing in the direction of the field. His wedding ring has a diameter of 2.23 cm, and it takes 0.320 s to move it into the field.
1. What average current is induced in the ring in A if its resistance is 0.0100 Ω?

2. What average power is dissipated in mW?

3. What magnetic field is induced at the center of the ring in T?
Physics
1 answer:
Ghella [55]3 years ago
7 0

Given that,

Magnetic field strength is

B = 1.8T

The wedding ring has a diameter of

d = 2.23 cm = 0.023m

Time take t = 0.32 secs

A. Current induced

From ohms law

V= iR, given that R = 0.01Ω

So, we need to get the induced emf

Using

ε = -NdΦ / dt

Where Φ = BA

ε = -A ∆B / ∆t

ε = -¼πd²(B2-B1) / (t2-t1)

ε = -¼ × π × 0.023² × -1.8 / 0.32

ε = 0.0023371 V

Then

I = ε / R

I = 0.002337 / 0.01

I =0.2337 A

B. Power discippated?

Power is given as

P = iV

P = 0.2337 × 0.002337

P = 0.0005462 W

P = 0.56 mW

C. The magnetic field at the centre of the ring.

The electric field at the centre of the ring is zero because each part of the ring will cause a symmetrical opposite magnitude at every point,

Then, B = 0 T

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Answer:

Explanation:

Average velocity is change in position over the change in time. You have to take the directions into account. 100km east and then 100km south. That forms a right triangle, so you can use Pythagorean's theorem to find the hypotenuse which would be the displacement from his initial position:

\sqrt{100^2+100^2} = \sqrt{20,000}=141.42km

Now take the displacement and divide it by the time it took to do the trip, 3 hours:

141.42km/3hours=47.14km/h

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Answer:

mechanical

Explanation:

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Which scientist discovered the laws that determine the motion of a spring?
Kruka [31]

Answer: Robert Hooke

Explanation: Hooke's law, law of elasticity discovered by the English scientist Robert Hooke in 1660, which states that, for relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load.

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2 years ago
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Annette [7]

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3 0
3 years ago
In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
3 years ago
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