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lidiya [134]
2 years ago
9

A closely wound search coil has an area of 3.13 cm2, 135 turns, and a resistance of 61.1 Ω. It is connected to a charge-measurin

g instrument whose resistance is 44.4 Ω. When the coil is rotated quickly from a position parallel to a uniform magnetic field to one perpendicular to the field, the instrument indicates a charge of 3.44×10−5 C . What is the magnitude of the magnetic field?
Physics
1 answer:
erastovalidia [21]2 years ago
8 0

Answer:

Explanation:

Let the magnitude of magnetic field be B .

flux passing through the coil's  = area of coil x field x no of turns

Φ = 3.13 x 10⁻⁴ x B x 135 = 422.55 x 10⁻⁴ B .

emf induced = dΦ / dt , Φ is magnetic flux.

current i = dΦ /dt x 1/R

charge through the coil = ∫ i dt

= ∫   dΦ /dt x 1/R dt

= 1 / R ∫ dΦ

= Φ / R

Total resistance R = 61.1 + 44.4 = 105.5 ohm .

3.44 x 10⁻⁵ = 422.55 x 10⁻⁴ B / 105.5

B = 3.44 x 10⁻⁵ x 105.5  / 422.55 x 10⁻⁴

= .86 x 10⁻¹

= .086 T .

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Answer:

-2.26×10^-4 radians

Explanation:

The solution involves a right angle triangle

Length is z while the horizontal is the height x

X^2+ 100^2=z^2

Taking the derivatives

2x(dx/dt)=Z^2(dz/dt)

Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11

dz/dt= 1100sqrt3/200 = 9.53

Sin a= 100/a

Taking derivatives in terms of t

Cos a(da/dt)=100/z^2 dz/dt

a= 30°

Cos (30°)da/dt= (-100/40000×9.5)

a= -2.26×10^-4radians

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3 years ago
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In a closed system, a cart with a mass of 1.5 kg is rolling to the right at 1.4 m/s, while another cart of mass 1.0 kg is rollin
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Answer: The correct option is (c.).

Explanation:

Mass of the cart A= 1.5 kg

Velocity of Cart A = 1.4 m/s towards right

Mass of the cart B = 1.0 kg

Velocity of Cart B = 1.4 m/s towards left

Momentum (P)= Mass × Velocity

P_A=1.5 kg\times 1.4 m/s=2.1 kg m/s

P_B=1.0 kg\times (-1.4m/s)=-1.4 kg m/s

(Negative sign means velocity of the cart is in opposite direction of that of the cart A)

Total Momentum =P_A+P_B=2.10 kg m/s-1.40 kg m/s=0.70 kg m/s

Hence, the correct option is (c.).

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3 years ago
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An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the sol
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Answer:

(a) B = 2.85 × 10^{-6} Tesla

(b) I =  I = 0.285 A

Explanation:

a. The strength of magnetic field, B, in a solenoid is determined by;

r = \frac{mv}{qB}

⇒ B = \frac{mv}{qr}

Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field

B = \frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }

  = \frac{9.11*10^{-27} }{3.2*10^{-21} }

B = 2.85 × 10^{-6} Tesla

b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

B = μ I N/L

⇒    I = B ÷ μN/L

where B is the magnetic field,  μ is the permeability of free space = 4.0 ×10^{-7}Tm/A, N/L is the number of turns per length.

I = B ÷ μN/L

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I = 0.285 A

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Please solve the Problem.
STatiana [176]

(a) For series circuit, current in 14 ohms = current in 72 ohms = 0.698 A

(b) Power loss in each series resistor, 14 ohms = 6.82 W and 72 ohms = 35.1 W.

(c) For parallel circuit, current in 14 ohms = 4.29 A and current in 72 ohms = 0.83 A

(d) Power loss in each parallel resistor, 14 ohms = 257.4 W and 72 ohms = 49.8 W.

<h3>Current in each series resistors</h3>

The total resistance = R1 + R2

                              R = 14 + 72 = 86 ohms

Current = V/R = 60/86 = 0.698 A

Since the resistors are in series, current in 14 ohms = current in 72 ohms = 0.698 A

<h3>Power loss in each series resistor</h3>

P = I²R

P(14 ohms) = (0.698)² x 14 = 6.82 W

P(72 ohms) = (0.698)² x 72 = 35.1 W

<h3>Current in each resistor for parallel arrangement</h3>

Total resistance, 1/R = 1/R1 + 1/R2

1/R = 1/14 + 1/72

1/R = 0.0853

R = 1/0.0853

R = 11.72

Total current in the circuit = V/R = 60/11.72 = 5.12 A

Current in 14 ohms = 60/14 = 4.29 A

Current in 72 ohms = 60/72 = 0.83 A

<h3>Power loss in each parallel resistor</h3>

P = IV

P(14 ohms) = (4.29) x 60 = 257.4 W

P(72 ohms) = 0.83 x 60 = 49.8 W

Learn more about current here: brainly.com/question/24858512

#SPJ1

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2 years ago
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