Two long thin,parallel conductors carrying eq | vedclass

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

Question

(D) The magnetic field due to a long straight wire carrying current $I$ at a distance $r$ is $B = \frac{\mu_0 I}{2 \pi r}$.$(A)$ On the $x$-axis,the d

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) The magnetic field due to a long straight wire carrying current $I$ at a distance $r$ is $B = \frac{\mu_0 I}{2 \pi r}$.$(A)$ On the $x$-axis,the distance from both wires is $r = a$. The magnetic fields are equal in magnitude and opposite in direction,so $B_{net} = 0$. This is correct.$(B)$ On the $y$-axis at $(0, y, 0)$,the distance to the wire at $y=a$ is $|a-y|$ and to the wire at $y=-a$ is $|a+y|$. Using the right-hand rule,the fields are in the $z$-direction. Thus,$B$ has only a $z$-component. This is correct.$(C)$ On the $z$-axis at $(0, 0, z)$,the distance to both wires is $r = \sqrt{a^2 + z^2}$. By symmetry,the $z$-components of the magnetic fields cancel out,and the $y$-components add up. Thus,$B$ has only a $y$-component. This is correct.Since all statements are correct,the answer is $(D)$.

Two long thin,parallel conductors carrying equal currents $I$ in the same direction are fixed parallel to the $x$-axis,one passing through $y = a$ and the other through $y = -a$. The resultant magnetic field due to the two conductors at any point is $B$. Which of the following are correct?

Similar Questions

The Biot-Savart law for a moving point charge $q$ with velocity $\vec{v}$ is given by $\vec{B} = \frac{\mu_0}{4\pi} \frac{q(\vec{v} \times \vec{r})}{r^3}$. This indicates that the magnetic field $\vec{B}$ produced by an electron moving with velocity $\vec{v}$ is such that:

$Weber/m^2$ is equal to

An electron of mass $m$ is revolving around the nucleus in a circular orbit of radius $r$ and has angular momentum $L$. The magnetic field produced by the electron at the centre of the orbit is ($e = \text{electric charge}$, $\mu_{0} = \text{permeability of free space}$)

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 $A$).

$A$ current of $8 \text{ A}$ each flows in opposite directions in two parallel conducting wires placed at a distance of $30 \text{ cm}$. The magnitude of the magnetic field at the midpoint between the two wires is . . . . . . $\mu \text{T}$. (Given: $\frac{\mu_0}{4\pi} = 10^{-7} \text{ N/A}^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module