Ask question

Ask question

All categories

2 years ago

12

8) A flat circular loop having one turn and radius 5.0 cm is positioned with its plane perpendicular to a uniform 0.60-T magneti

c field. The area of the loop is suddenly reduced to essentially zero in 0.50 ms. What emf is induced in the loop

1 answer:

Marrrta [24]2 years ago

7 0

Answer:

EMF induced in the loop is 9.4 V

Explanation:

As we know that initial magnetic flux of the loop is given as

$\phi_1 = B.A$

$\phi_1 = (0.60)(\pi (0.05)^2)$

$\phi_1 = 4.7 \times 10^{-3} Wb$

As soon as the area of the loop becomes zero the final magnetic flux of the loop is ZERO

Now as per faraday's law of electromagnetic induction the EMF is induced due to rate of change in magnetic flux

so we will have

$EMF = \frac{\Delta \phi}{\Delta t}$

so we will have

$EMF = \frac{4.7 \times 10^{-3} - 0}{0.50 \times 10^{-3}}$

$EMF = 9.4 V$

You might be interested in

You go to an amusement park with your friend Betty, who wants to ride the 70-m-diameter Ferris wheel. She starts the ride at the

Ksenya-84 [330]

Answer:

Part a)

$d = 23.94 m$

Part b)

$\theta = 110 degree$

Explanation:

Part a)

As we know that initially the position vector is r

then the same magnitude position vector is rotated by 40 degree angle

so displacement magnitude is the magnitude of change in position vector

so it is given as

$d = \sqrt{r_1^2 + r_2^2 + 2r_1r_2cos(180-\theta)}$

$r_1 = r_2 = 35 m$

$d = \sqrt{35^2 + 35^2 -2(35)(35)cos40}$

$d = 23.94 m$

Part b)

now we need to find the direction of the displacement vector

so let say it makes an angle with x axis so we have

$tan\theta = \frac{rsin\phi}{r - rcos\phi}$

$tan\theta = \frac{sin 40}{1 - cos40}$

$\theta = 110 degree$

4 0

2 years ago

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows. The pulley can be tre

tankabanditka [31]

Answer:

Mass of the pull is 77 kg

Explanation:

Here we have for

Since the rope moves along with pulley, we have

For the first block we have

T₁ - m₁g = -m₁a = -m₁g/4

T₁ = 3/4(m₁g) = 323.4 N

Similarly, as the acceleration of the second block is the same as the first block but in opposite direction, we have

T₂ - m₂g = m₂a = m₂g/4

T₂ = 5/4(m₂g) = 134.75 N

T₂r - T₁r = I·∝ = 0.5·M·r²(-α/r)

∴ $M = -\frac{2}{a} (T_2-T_1)$

$M = -\frac{2}{2.45} (134.75-323.4) = 77 \, kg$

Mass of the pull = 77 kg.

5 0

2 years ago

A propagating wave in space with electric and magnetic components. These components oscillate at right angles to each other. It

zimovet [89]

Answer:

electromagnetic

Explanation:

8 0

2 years ago

Read 2 more answers

The pfsense firewall, like other firewalls on the market, relies on __________ to expose an ip address from the private network

sashaice [31]

T<span>he pfsense firewall, like other firewalls on the market, relies on the subnet mask to expose an ip address from the private network and bind it to an address on the public network. </span>

6 0

2 years ago

A long, straight wire carrying a current of 3.45 A moves with a constant speed v to the right. A 5-turn circular coil of diamete

d1i1m1o1n [39]

Answer:

I = 69.3 μA

Explanation:

Current through the straight wire, I = 3.45 A

Number of turns, N = 5 turns

Diameter of the coil, D = 1.25 cm

Resistance of the coil, $R = 3.25 \mu ohms$

Distance of the wire from the center of the coil, d = 20 cm = 0.2 m

The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.

$B_{1} = \frac{\mu_{0}I }{2\pi d}$

$B_{1} = \frac{4\pi* 10^{-7} *3.45 }{2\pi *0.2}\\B_{1} =0.00000345 T$

Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil

$B_{2} = \frac{\mu_{0}I }{2\pi(2d)) } \\B_{2} = \frac{\mu_{0}I }{4\pi d } \\$

$B_{2} = \frac{4\pi* 10^{-7} *3.45 }{2\pi *2*0.2}\\B_{2} = 0.000001725 T$

Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345

ΔB = -0.000001725

Induced current, $I = \frac{E}{R}$

E = -N (Δ∅)/Δt

Δ∅ = A ΔB

Area, A = πr²

diameter, d = 0.0125 m

Radius, r = 0.00625 m

A = π* 0.00625²

A = 0.0001227 m²

Δ∅ = -0.000001725 * 0.0001227

Δ∅ = -211.6575 * 10⁻¹²

E = -N (Δ∅)/Δt

$E = -5\frac{-211.6575 * 10^{-12} }{4.70} \\E = 225.17 * 10^{-12} V$

Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms

I = E/R

$I = \frac{225.17 * 10^{-12} }{3.25 * 10^{-6} }$

I = 0.0000693 A

I = 69 .3 * 10⁻⁶A

I = 69.3 μA

3 0

2 years ago