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Marina86 [1]
2 years ago
11

A 2 m long piece wire of attached to a 3 V battery has a magnetic force of 8 N acting on it as it

Physics
2 answers:
Reptile [31]2 years ago
6 0

Answer:

<h2>R = 3.6Ω</h2>

Explanation:

Before we can calculate the resistance of the wire, we need to first calculate the Force experienced by the wire in the magnetic field. The force experienced by  the wire is expressed as;

F = BIL

B is the magnetic field measured in Tesla = 4.8T

I is the current in the wire

L is the length of the wire = 2m

F is magnetic force acting on the wire = 8N

from the formula I = F/BL

I = 8/4.8*2

I = 8/9.6

I = 0.83Amperes

According to ohms law;

V=IR

R = V/I

given Voltage = 3V

R = 3/0.83

R = 3.6Ω

The resistance of the wire to nearest tenth is 3.6Ω to nearest tenth

Ira Lisetskai [31]2 years ago
4 0

Answer:

The resistance of the wire is 3.6 Ohm's.

Explanation:

The force on a conductor in a magnetic field is given as;

         F = BILSinθ

where: F is the force, i is the current through the conductor, L is the length of the conductor and θ is the angle between the conductor and field.

From the question, F = 8N, B = 4.8T, L = 2m, I = ? and θ = 90^{0}.

So that,

    8 = 4.8 × I × 2 × Sin90^{0}

    8 = 9.6I

    I = \frac{8}{9.6}

    I = 0.8333A

The current flowing through the wire is 0.83A.

to determine the resistance of the wire, Ohm's law states that:

     V = IR

where v is the electromotive force, I is the current flowing and R is the resistance of the conductor.

V = 3V, I = 0.8333, then;

   R = \frac{V}{I}

      = \frac{3}{0.8333}

     = 3.6001Ohm's

The resistance of the wire is 3.6 Ohm's.

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The RMS potential difference of an AC household outlet is 117 V. a) What is the maximum potential difference across a lamp conne
Anit [1.1K]

Answer:

a. 165.5 V

b. 7.78 A

Explanation:

Here is the complete question

The RMS potential difference of an AC household outlet is 117 V. a) What is the maximum potential difference across a lamp connected to the outlet? b) If the RMS current through the lamp is 5.5 A, what is the maximun current through the lamp.

Solution

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V₀ = √2V₁ = √2 × 5.5 A = 7.78 A

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