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svlad2 [7]
3 years ago
8

A circular loop of wire with radius 2.00 cm and resistance 0.600 Ω is in a region of a spatially uniform magnetic field B⃗ that

is perpendicular to the plane of the loop. At t = 0 the magnetic field has magnitude B0=3.00T. The magnetic field then decreases according to the equation B(t)=B0e−t/τ, where τ=0.500s.
Physics
1 answer:
Anika [276]3 years ago
5 0

Answer:

Incomplete questions

Let assume we are asked to find

Calculate the induced emf in the coil at any time, let say t=2

And induced current

Explanation:

Flux is given as

Φ=NAB

Where

N is number of turn, N=1

A=area=πr²

Since r=2cm=0.02

A=π(0.02)²=0.001257m²

B=magnetic field

B(t)=Bo•e−t/τ,

Where Bo=3T

τ=0.5s

B(t)=3e(−t/0.5)

B(t)=3exp(-2t)

Therefore

Φ=NAB

Φ=0.001257×3•exp(-2t)

Φ=0.00377exp(-2t)

Now,

Induce emf is given as

E= - dΦ/dt

E= - 0.00377×-2 exp(-2t)

E=0.00754exp(-2t)

At t=2

E=0.00754exp(-4)

E=0.000138V

E=0.138mV

b. Induce current

From ohms laws

V=iR

Given that R=0.6Ω

i=V/R

i=0.000138/0.6

i=0.00023A

i=0.23mA

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5 0
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Can someone help me with this. I'm not really sure if the right answer is c.
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It might be c I'm not sure either. Hope I helped
6 0
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Weight is a force that depends on mass and ?
Maksim231197 [3]

Answer:

Gravitational force.

8 0
3 years ago
A manometer is used to measure the air pressure in a tank. the fluid used has a specific gravity of 1.25, and the differential h
BartSMP [9]
Specific Gravity of the fluid = 1.25 
Height h = 28 in
 Atmospheric Pressure = 12.7 psia
 Density of water = 62.4 lbm/ft^3 at 32F
 Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
 Density of the Fluid p = 78 lbm/ft^3
 Difference in pressure as we got the differential height, dP = p x g x h  dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
 Difference in pressure = 1.26 psia
 (a) Pressure in the arm that is at Higher 
 P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
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4 0
4 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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