Question 48: Magnetism and Matter

Question

A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at $$\frac{a}{2}$$ and $$2 a$$ from axis of the wire is :

Accepted Answer

To find the ratio of the magnetic field at $$\frac{a}{2}$$ and $$2a$$ distances from the axis of a long straight wire, we use AmpÃÂ¨re's Law, which relates the magnetic field around a current-carrying conductor to the current enclosed by it.
AmpÃÂ¨reÃ¢ÂÂs Law is given by:
$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{	ext{enc}}$$
where:
$$\vec{B}$$ is the magnetic field,
$$\mu_0$$ is the permeability of free space,
$$I_{	ext{enc}}$$ is the enclosed current.
For a point inside the wire (at radius $$r=a/2$$):
The current enclosed by a radius $$r$$ is proportional to the area of the cross-section at radius $$r$$.
The area of the cross-section at radius $$r$$ is given by:
$$\pi \left( \frac{a}{2} ight)^2 = \frac{\pi a^2}{4}$$
The total current $$I$$ is uniformly distributed, thus the current enclosed $$I_{	ext{enc}}$$ at radius $$r = \frac{a}{2}$$ is:
$$I_{	ext{enc}} = I 	imes \frac{	ext{Area enclosed}}{	ext{Total area}} = I 	imes \frac{\frac{\pi a^2}{4}}{\pi a^2} = \frac{I}{4}$$
Applying AmpÃÂ¨reÃ¢ÂÂs Law inside the conductor, we get:
$$B \cdot 2 \pi \left( \frac{a}{2} ight) = \mu_0 \left( \frac{I}{4} ight)$$
So,
$$B \cdot \pi a = \frac{\mu_0 I}{4}$$
Therefore, the magnetic field inside the wire at $$r = \frac{a}{2}$$ is:
$$B_{\frac{a}{2}} = \frac{\mu_0 I}{4 \pi a}$$
For a point outside the wire (at radius $$r = 2a$$):
The total current enclosed by a radius $$r = 2a$$ is the entire current $$I$$.
Applying AmpÃÂ¨reÃ¢ÂÂs Law outside the conductor, we get:
$$B \cdot 2 \pi (2a) = \mu_0 I$$
So,
$$B \cdot 4 \pi a = \mu_0 I$$
Therefore, the magnetic field outside the wire at $$r = 2a$$ is:
$$B_{2a} = \frac{\mu_0 I}{4 \pi a}$$
Hence, the ratio of the magnetic field at $$\frac{a}{2}$$ and $$2a$$ is:
$$\frac{B_{\frac{a}{2}}}{B_{2a}} = \frac{\frac{\mu_0 I}{4 \pi a}}{\frac{\mu_0 I}{4 \pi a}} = 1:1$$
So, the correct option is:
Option C: $$1:1$$

Answer

$$4: 1$$

Answer

$$3: 4$$

Answer

$$1: 1$$

Answer

$$1: 4$$

Magnetism and Matter

A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at $\frac{a}{2} and 2 a$ from axis of the wire is :

Choose the correct answer: