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REY [17]
3 years ago
14

The plane of a rectangular coil, 7.2 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil

has 104 turns and a total resistance of 12.4 Ω, at what rate must the magnitude of B change to induce a current of 0.05 A in the windings of the coil? Answer in units of T/s.
Physics
1 answer:
netineya [11]3 years ago
4 0

Answer:

The rate of change of magnetic field is 2.23 T/s.              

Explanation:

Given that,

Dimension of rectangular coil is 7.2 cm by 3.7 cm.

Number of turns in the coil, N = 104

Resistance of the coil, R = 12.4 ohms

Current, I = 0.05 A

We need to find the rate of change of magnetic field in the coil. The induced emf is given by the rate of change of magnetic flux. So,

\epsilon=-\dfrac{d\phi}{dt}

Ohm's law is :

\epsilon=IR

So,

IR=-\dfrac{d\phi}{dt}\\\\IR=-\dfrac{d(NBA)}{dt}\\\\IR=-NA\dfrac{dB}{dt}\\\\\dfrac{dB}{dt}=\dfrac{IR}{NA}\\\\\dfrac{dB}{dt}=\dfrac{0.05\times 12.4}{104\times 7.2\times 10^{-2}\times 3.7\times 10^{-2}}\\\\\dfrac{dB}{dt}=2.23\ T/s

So, the rate of change of magnetic field is 2.23 T/s.

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Explanation:

For a charge concentrated nearly at a point, the electric field is directly proportional to the amount of charge; it is inversely proportional to the square of the distance radially away from the centre of the source charge and depends also upon the nature of the medium.

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3 years ago
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One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a
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This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

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3 years ago
Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat
nikklg [1K]

Answers:

a) T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds V are given by:

V=\omega L=\frac{2\pi}{T} L (1)

Where:

\omega is the angular frequency

L is the radius of the orbit of each satellite

T is the period of the orbit of each satellite

Isolating T:

T=\frac{2 \pi L}{V} (2)

Applying this equation to each satellite:

T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s (3)

T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s (4)

T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s (5)

T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s (6)

T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s (7)

T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s (8)

Ordering this periods from largest to smallest:

T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) Acceleration a is defined as the variation of velocity in time:

a=\frac{V}{T} (9)

Applying this equation to each satellite:

a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2} (10)

a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2} (11)

a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2} (12)

a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2} (13)

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a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2} (15)

Ordering this acceerations from largest to smallest:

a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

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This question involves the concepts of average speed, distance, and time.

The average speed of the family is "2.5 mi/hr".

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The average speed is defined as the ratio of the total distance covered with the total time taken to cover it.

v=\frac{s}{t}

where,

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Therefore,

v=\frac{2.5\ mi}{1\ hr}

v = 2.5 mi/hr

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Answer:  1.3439 x 10⁵ (days²/AU³)

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