Question

The magnetic field 10 cm from a wire carrying a 1 A current is 2 μT. What is the field 4 cm from the wire? Express your answer with the appropriate units.

Answer 1

Answer:

$5 \mu T$

Explanation:

We can solve the problem in two ways:

1) The magnetic field around a current-carrying wire is given by

$B=\frac{\mu_0 I}{2 \pi r}$

where

$\mu_0 = 4\pi \cdot 10^{-7}Tm/A$ is the permeabilty of free space

I is the current

r is the distance from the wire

Substituting I=1 A and $r=4 cm=0.04 m$, we find:

$B=\frac{(4\pi \cdot 10^{-7})(1 A)}{2 \pi (0.04 m)}=5\cdot 10^{-6} T=5 \mu T$

2) We can notice that the magnitude of the magnetic field around a current-carrying wire is inversely proportional to the distance from the wire:

$B\propto \frac{1}{r}$

which means that the product $B \cdot r$ is constant if we keep the same wire. So we can write

$B_1 r_1 = B_2 r_2$

and using $B_1 = 2 \mu T, r_1 = 10 cm, r_2 = 4 cm$ we can find the magnetic field at 4 cm from the wire:

$B_2 = \frac{B_1 r_1}{r_2}=\frac{(2 \mu T)(10 cm)}{4 cm}=5 \mu T$

Answer 2

The field 4 cm from the wire is 5 μT

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Further explanation

Let's recall magnetic field strength from current carrying wire:

$\large {\boxed {B = \mu_o \frac{I}{2 \pi d} } }$

B = magnetic field strength (T)

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

I = current (A)

d = distance (m)

Let's tackle the problem now !

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Given:

initial distance = d₁ = 10 cm

current = I = 1 A

initial magnetic field strength = B₁ = 2 μT

final distance = d₂ = 4 cm

Asked:

final magnetic field strength = B₂ = ?

Solution:

$B_1 : B_2 = \mu_o \frac{I}{2\pi d_1} : \mu_o \frac{I}{2\pi d_2}$

$B_1 : B_2 = \frac{1}{d_1} : \frac{1}{d_2}$

$B_1 : B_2 = d_2 : d_1$

$B_2 = \frac{d_1}{d_2} \times B_1$

$B_2 = \frac{10}{4} \times 2$

$B_2 = 5 \ \mu T$

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Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Magnetic Field

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Keywords: Magnet , Field , Magnetic , Current , Wire , Unit ,