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Elan Coil [88]
3 years ago
5

A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f

rom a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is B. Part A What is the magnitude Φinitial of the magnetic flux through one turn of the coil before it is rotated?
Physics
1 answer:
anzhelika [568]3 years ago
6 0

Answer:

\phi_i = BA

Explanation:

magnetic flux is the count of magnetic field lines passing through a given loop or area

As we know that magnetic flux is given by the formula

\phi = \vec B. \vec A

here we also know that magnetic field B and plane of the coil is perpendicular in initial position

So the area vector is always perpendicular to the plane of the coil

so the angle between magnetic field and area vector is parallel to each other and this angle would be zero

so magnetic flux of the coil initially we have

\phi = BAcos0 = BA

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A circuit has resistors R 1 = 468 Ω and R 2 = 125 Ω R1=468 Ω and R2=125 Ω , and two batteries V 1 = 12.0 V and V 2 = 3.00 V V1=1
Greeley [361]

Answer:

0.0192A

Explanation:

Since, the reading of the galvanometer is 0 A, the voltage across the resistance R will be:

Step 1

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VR = 3.00v

Step 2

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IR = V1 - V2 /R1

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To solve this problem we will apply Ohm's law. The law establishes that the potential difference V that we apply between the ends of a given conductor is proportional to the intensity of the current I flowing through the said conductor. Ohm completed the law by introducing the notion of electrical resistance R. Mathematically it can be described as

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3 0
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Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

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  • final velocity of the sprinter, v = 27 km/h
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Convert the velocity of the sprinter to m/s;

initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\

The time of motion is seconds;

t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s

The sprinter’s average acceleration is calculated as follows;

a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2

Thus, the sprinter’s average acceleration is 1.98 m/s²

Learn more here:brainly.com/question/17280180

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