A straight wire of length pi^2 , m carries a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. For a straight wire of length $L = \pi^2 \, m$ carrying current $i = 2 \, A$,the magnetic field $B_1$ at a distance $d = 1 \, cm = 10^{-2} \,

Vedclass · Accepted Answer

$1:50$

Vedclass · Answer

(B) $1$. For a straight wire of length $L = \pi^2 \, m$ carrying current $i = 2 \, A$,the magnetic field $B_1$ at a distance $d = 1 \, cm = 10^{-2} \, m$ is given by the Biot-Savart law for an infinitely long wire (assuming the point is near the center of the wire): $B_1 = \frac{\mu_0 i}{2\pi d} = \frac{\mu_0 	imes 2}{2\pi 	imes 10^{-2}} = \frac{\mu_0}{\pi 	imes 10^{-2}}$.$2$. When the wire is bent into a circle of radius $R$,its circumference is $2\pi R = L = \pi^2$. Thus,$R = \frac{\pi^2}{2\pi} = \frac{\pi}{2} \, m$.$3$. The magnetic field at the center of this circular loop is $B_2 = \frac{\mu_0 i}{2R} = \frac{\mu_0 	imes 2}{2 	imes (\pi/2)} = \frac{\mu_0}{\pi/2} = \frac{2\mu_0}{\pi}$.$4$. The ratio of the magnetic field at the center to the first case is $\frac{B_2}{B_1} = \frac{2\mu_0 / \pi}{\mu_0 / (\pi 	imes 10^{-2})} = 2 	imes 10^{-2} = \frac{2}{100} = \frac{1}{50}$.$5$. Therefore,the ratio is $1:50$.

Similar Questions

In the figure,two parallel infinitely long current-carrying wires are shown. If the resultant magnetic field at point $A$ is zero,then determine the current $I$ (in $A$).

Equal current $i$ is flowing in three infinitely long wires along positive $x, y$ and $z$ directions. The magnetic field at a point $(0, 0, -a)$ would be:

Write the formula for the magnetic field at a point on the axis of a circular current-carrying loop of radius $R$ at a distance $x$ from its center,where $x >> R$.

The unit of permeability of vacuum $(\mu_{0})$ is . . . . . . .

$A$ steady current is set up in a cubic network composed of wires of equal resistance and length $d$ as shown in the figure. What is the magnetic field at the centre $P$ due to the cubic network?

Vedclass Test Series

Exam Paper Generator

Online Exam Module