Question

A long straight wire (radius = 3.0 mm) carries a constant current distributed uniformly over a cross section perpendicular to the axis of the wire. If the current density is 100 A/m². The magnitudes of the magnetic field at (a) 2.0 mm from the axis of the wire and (b) 4.0 mm from the axis of the wire is :-

• A. 2πx10⁻⁸T, $\frac{9\pi}{2}$x10⁻⁸T
• B. 4π × 10⁻⁸ T, $\frac{\pi}{2}$x10⁻⁸T
• C. 4π×10⁻⁸T, $\frac{9\pi}{2}$×10⁻⁸T
• D. π×10⁻⁸ T, $\frac{9\pi}{2}$×10⁻⁸ T

Answer 1

Answer: (C)

(3)

Answer 2

To determine the magnetic field at points inside and outside a long straight wire with a uniform current density, we use Ampere's Law.

Given:

• Radius of the wire, $R = 3.0 \text{ mm} = 3.0 \times 10^{-3} \text{ m}$

• Current density, $J = 100 \text{ A/m}^2$

• Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$

Formulas for Magnetic Field ($B$) due to a Long Straight Wire with Uniform Current Density ($J$):

1. Inside the wire ($r < R$):
   For an Amperian loop of radius $r$ inside the wire, the enclosed current $I_{enclosed}$ is the current density times the area of the loop:
   $I_{enclosed} = J \times (\pi r^2)$
   Applying Ampere's Law ($\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$):
   $B_{inside} (2\pi r) = \mu_0 (J \pi r^2)$
   $B_{inside} = \frac{\mu_0 J r}{2}$

2. Outside the wire ($r > R$):
   For an Amperian loop of radius $r$ outside the wire, the enclosed current $I_{enclosed}$ is the total current in the wire, $I_{total}$.
   $I_{total} = J \times (\pi R^2)$
   Applying Ampere's Law:
   $B_{outside} (2\pi r) = \mu_0 I_{total}$
   $B_{outside} (2\pi r) = \mu_0 (J \pi R^2)$
   $B_{outside} = \frac{\mu_0 J R^2}{2r}$

Calculations:

(a) Magnetic field at $r_a = 2.0 \text{ mm}$ from the axis of the wire:
Since $r_a = 2.0 \text{ mm} < R = 3.0 \text{ mm}$, this point is inside the wire.
Using the formula for $B_{inside}$:
$B_a = \frac{\mu_0 J r_a}{2}$
$B_a = \frac{(4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}) \times (100 \text{ A/m}^2) \times (2.0 \times 10^{-3} \text{ m})}{2}$
$B_a = \frac{4\pi \times 10^{-7} \times 10^2 \times 2 \times 10^{-3}}{2} \text{ T}$
$B_a = 4\pi \times 10^{-7} \times 10^{-1} \text{ T}$
$B_a = 4\pi \times 10^{-8} \text{ T}$

(b) Magnetic field at $r_b = 4.0 \text{ mm}$ from the axis of the wire:
Since $r_b = 4.0 \text{ mm} > R = 3.0 \text{ mm}$, this point is outside the wire.
Using the formula for $B_{outside}$:
$B_b = \frac{\mu_0 J R^2}{2r_b}$
$B_b = \frac{(4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}) \times (100 \text{ A/m}^2) \times (3.0 \times 10^{-3} \text{ m})^2}{2 \times (4.0 \times 10^{-3} \text{ m})}$
$B_b = \frac{4\pi \times 10^{-7} \times 10^2 \times (9.0 \times 10^{-6})}{8.0 \times 10^{-3}} \text{ T}$
$B_b = \frac{36\pi \times 10^{-11}}{8 \times 10^{-3}} \text{ T}$
$B_b = \frac{36\pi}{8} \times 10^{-8} \text{ T}$
$B_b = \frac{9\pi}{2} \times 10^{-8} \text{ T}$

Comparing these results with the given options, we find that:
(a) $B_a = 4\pi \times 10^{-8} \text{ T}$
(b) $B_b = \frac{9\pi}{2} \times 10^{-8} \text{ T}$

These values match option (3).

The final answer is $\boxed{\text{(3)}}$

Solution Explanation:
The problem requires calculating the magnetic field at two points, one inside and one outside a long straight wire with uniform current density. Ampere's Law is applied for both cases.

1. Inside the wire ($r < R$): The magnetic field is directly proportional to the distance $r$ from the axis, given by $B = \frac{\mu_0 J r}{2}$.

2. Outside the wire ($r > R$): The magnetic field is inversely proportional to the distance $r$ from the axis, and proportional to the square of the wire's radius, given by $B = \frac{\mu_0 J R^2}{2r}$.

Substitute the given values for current density, wire radius, and distances to find the magnetic field magnitudes at the specified points.