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Elenna [48]
2 years ago
10

A 100-turn coil has a radius of 7.50 cm and a resistance of 50.0 W. At what rate must a perpendicular magnetic field change to p

roduce a current of 5.00 A in the coil
Physics
1 answer:
Furkat [3]2 years ago
3 0

Answer:

dB / dt = -141.47 T / s

Explanation:

For this exercise let's use Faraday's law to calculate the induced electromotive force

          E = - N dФ / dt

the magnetic flux is

         Ф = B. A

in this case the direction of the field and the normal to area are parallel so the scalar product is reduced to the algebraic product

         Ф = B A

  we substitute

        E = - N A dB / dt

the area of ​​the loop is

        A = π r²

we substitute

          E = - N π r² dB / dt

in the exercise indicate that the resistance of the coil is R = 50.0 Ω,

           E = i R

we substitute

           i R = -N π r² dB / dt

           dB / dT = - i R / N π r²

let's calculate

           dB / dt = - 5.00 50.0 / (100 π 0.075²)

           dB / dt = -141.47 T / s

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3 years ago
A Honda Hawk motorcycle and its rider with a combined mass of 450 kg travels around a curve of radius 106 m with a speed of 18 m
umka21 [38]

Answer: coefficient of static friction

= 0.31

Explanation: Since they negotiate the curve without skidding, the frictional force (F1) equals the centripetal force (F2).

F1= uN

F2 = M*(v²/r)

M is the combined mass 450kg

V is the velocity 18m/s

r is the radius 106m

N is the normal reaction 4410N

u is the coefficient of static friction

Making u subject of the formula we have that,

u = {450*(18²/106)} /4410

=1375.47/4410

=0.31

NOTE: coefficient of friction is dimensionless. It as no Unit.

7 0
2 years ago
A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

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