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Anastaziya [24]
3 years ago
13

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant

rate of 12.0cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop of magnitude 0.500T. Find the emf induced in the loop, at the instant when 9.0s have passed. And find the direction

Physics
1 answer:
vovangra [49]3 years ago
6 0

Answer:

emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.

Direction is counter clockwise.

Explanation:

See attached pictures.

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Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in
Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

L = \frac{N^2 \mu_0 A}{l}

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2

L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH

(b) The energy stored in the inductor when 21 A current ;

E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s

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Answer:

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Lets take

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a'=\dfrac{F}{M}

Here given that M > m that is why a > a'

Therefore car will experiences more acceleration.

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