Jumper cables used to start a stalled vehicle often carry a 65 a current. how strong is the magnetic field 12 cm from one cable? t compare to the earth's magnetic field (5.5 ✕ 10-5 t).

Respuesta :

The magnetic field generated by a wire that carries a current of intensity I is :
[tex]B= \frac{\mu _0 I}{2 \pi r} [/tex]
where
[tex]\mu_0 = 4 \pi \cdot 10^{-7} T m A^{-1}[/tex] is the magnetic permittivity
I is the current
r is the radial distance from the wire at which the field is calculated.

In our problem, I=65 A and r=12 cm=0.12 m, so the magnetic field is:
[tex]B= \frac{(4 \pi \cdot 10^{-7} T m A^{-1})(65 A)}{(2 \pi)(0.12 m)}=1.1 \cdot 10^{-4} T [/tex]

And we can see that the intensity of this magnetic field at r=12 cm is approximately twice the intensity of the Earth's magnetic field ([tex]5.5 \cdot 10^{-5}T[/tex])

The magnetic field strength of the cable is about 1.1 × 10⁻⁴ T.

The magnetic field strength of the cable is twice the earth's magnetic field strength.

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Further explanation

Let's recall magnetic field strength from current carrying wire and from center of the solenoid as follows:

[tex]\boxed {B = \mu_o \frac{I}{2 \pi d} } [/tex]

B = magnetic field strength from current carrying wire (T)

μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)

I = current (A)

d = distance (m)

[tex]\texttt{ }[/tex]

[tex]\boxed {B = \mu_o \frac{I N}{L} } [/tex]

B = magnetic field strength at the center of the solenoid (T)

μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)

I = current (A)

N = number of turns

L = length of solenoid (m)

Let's tackle the problem now !

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Given:

Current = I = 65 A

Distance = d = 12 cm = 0.12 m

Permeability of free space = μo = 4π × 10⁻⁷ T.m/A

Earth's Magnetic Field Strength = Be = 5.5 × 10⁻⁵ T

Asked:

Cable's Magnetic Field Strength = B = ?

Solution:

We will use this folllowing formula to solve the problem:

[tex]B = \mu_o \frac{I}{2 \pi d}[/tex]

[tex]B = 4\pi \times 10^{-7} \times \frac{65}{2 \pi \times 0.12}[/tex]

[tex]\boxed{B \approx 1.1 \times 10^{-4} \texttt{ T}}[/tex]

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Next, we could compare this result with the earth's magnetic field as follows:

[tex]B : B_e = (1.1 \times 10^{-4}) : (5.5 \times 10^{-5})[/tex]

[tex]B : B_e = 2 : 1[/tex]

[tex]\boxed{B = 2B_e}[/tex]

[tex]\texttt{ }[/tex]

Learn more

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  • The three resistors : https://brainly.com/question/9503202
  • A series circuit : https://brainly.com/question/1518810
  • Compare and contrast a series and parallel circuit : https://brainly.com/question/539204

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Answer details

Grade: High School

Subject: Physics

Chapter: Magnetic Field

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