Formulae for determining the magnitude of the load overcomed

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

c)

In conclusion, the direction of the induced current will be Counterclockwise.

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4 0

2 years ago

ANSWER NOW!!!! IF CAN

liberstina [14]

Is this practically possible? How can a 100kg man fly? Hahaha

3 0

3 years ago

What is the force if the mass is 75kg and the acceleration is 24.5m/s^2

Snowcat [4.5K]

<u>Given;</u>

mass m = 75 kg

acceleration a = 24.5 ms²

<em>F = ma </em>

F = 75 kg * 24.5 ms²

= 1837.5 kg ms².

4 0

3 years ago

Please help me TT. I need this to be submitted soon TT. Thank you

Verizon [17]

Answer:

1)

a) f = 1m × 2 × (5A / √2) × (5A / √2) / 0.003m = 0.00166... (66 is repeating)

b) The currents on two wires on a AC chord are always moving in opposite direction and so they are always replusing.

c) There needs to be a sheath to dampen the replusing, fluctuating force of the wires.

2)

a) v = √( ( (-2)(-1.6 × 10^(-16))(3000V) ) / (2.84 × 10^(-20)kg) ) = 5.81227 × 10^3

b) Any ion transversing a chamber having a magnetic field will deflect.

c) The direction of the electric field is vertical because it's perpendicular to the plates. The electric field magnitude is independent from the magnitude of the magnetic field and charge. So it's not possible to find the magnitude of the electric field, without knowing the voltage on the plates, the distance between the plates, and the dielectric constant.

d) Assuming the mangetic field remained, the path of the negative ions will be deflected vertically given that the magnetic field is horizontally perpendicular to the negative charged ions movement.

Sorry it took so long :) If anything is incorrect please let me know.

3 0

3 years ago

PLEASE HELP !! A 10 ohm resistor is connected to a 9V battery how much current is flowing through the circuit?

kompoz [17]

Resitance (R)= 10 Ohm

Potential difference (V) = 9V

V= IR

I= V/R

I= 9/10

I= 0.9 Ampere

Therefore 0.9 Ampere of current is flowing through the circuit.

8 0

3 years ago

Formulae for determining the magnitude of the load overcomed​

Formulae for determining the magnitude of the load overcomed