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dexar [7]
3 years ago
14

Credit-Card Magnetic Strips Experiments carried out on the television show Mythbusters determined that a magnetic field of 1000

gauss is needed to corrupt the information on a credit card's magnetic strip. (They also busted the myth that a credit card can be demagnetized by an electric eel or an eelskin wallet.) Suppose a long, straight wire carries a current of 7.0A . How close can a credit card be held to this wire without damaging its magnetic strip?
Physics
1 answer:
Readme [11.4K]3 years ago
4 0

Answer:

14 μm

Explanation:

The magnetic field due to a long straight wire is B = μ₀i/2πr where  μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current = 7.0 A and r = distance of credit card from magnetic field.

So r = μ₀i/2πB since B = 1000 gauss = 1000 G × 1 T/10000 G  = 0.1 T

r = 4π × 10⁻⁷ H/m × 7.0 A/(2π × 0.1 T)

r = 2 × 10⁻⁷ H/m × 7.0 A/0.1 T

r = 14 × 10⁻⁷ H/m × A/0.1 T

r = 140 × 10⁻⁷ m

r = 1.4 × 10⁻⁵ m

r = 14 × 10⁻⁶ m

r = 14 μm

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3 years ago
A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm
I am Lyosha [343]

Answer:

Rate of change of magnetic field is 3.466\times 10^3T/sec        

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop r=\frac{0.13}{2}=0.065m

Length of the circular loop L=2\pi r=2\times 3.14\times 0.065=0.4082m

Wire is made up of diameter of 2.6 mm

So radius r=\frac{2.6}{2}=1.3mm=0.0013m

Cross sectional area of wire A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2

Resistivity of wire \rho =2.18\times 10^{-8}m

Resistance of wire R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm

Current is given i = 11 A

So emf  e=11\times 1.67\times 10^{-3}=0.0183volt

Emf induced in the coil is e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}

0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}

\frac{dB}{dt}=3.466\times 10^3=T/sec

8 0
3 years ago
A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi
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Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

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F = mx + b  ------ ( 1 )

m = slope  = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8   ----- ( 2 )

hence to determine the work done by the force

W   = \int\limits^7_0 { ( -0.373) x + 2.8  )} \, dx     Note:  the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

<u>b) calculate the speed of the train at the end of its journey</u>

we will apply the work energy theorem

W = 1/2 m*v^2  -  1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 )  ( input values into equation )

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hence V = 3.48 m/s

6 0
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Explanation:

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Explanation:

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F=G\frac{Mm}{r^2}

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G is the gravitational constant

M is the mass of the Sun

m is the mass of the planet

r is the distance between the Sun and the planet

This force acts as centripetal force, continuously "pulling" the planet towards the centre of its circular orbit.

2) The inertia of the planet. In fact, according to Newton's first law, an object in motion at constant velocity will continue moving at its velocity, unless acted upon an external unbalanced force. Therefore, the planet tends to continue its motion in a straight line (tangential to the circular orbit), however it turns in a circle due to the presence of the gravitational attraction of the Sun.

Learn more about gravity:

brainly.com/question/1724648

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