Two parallel conducting wires of equal length | vedclass

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

Question

(C) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.Since the curr

Vedclass · Accepted Answer

$\frac{\mu_0(I_1+I_2)}{\pi d}$

Vedclass · Answer

(C) The magnetic field $B$ produced by a long straight current-carrying wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.Since the currents $I_1$ and $I_2$ are in opposite directions,by the Right-Hand Thumb Rule,the magnetic fields produced by both wires at the midpoint will point in the same direction.Therefore,the resultant magnetic field is the sum of the individual fields: $B_{net} = B_1 + B_2$.The distance of the midpoint from each wire is $r = \frac{d}{2}$.Substituting the values: $B_{net} = \frac{\mu_0 I_1}{2\pi(d/2)} + \frac{\mu_0 I_2}{2\pi(d/2)}$.Simplifying the expression: $B_{net} = \frac{\mu_0 I_1}{\pi d} + \frac{\mu_0 I_2}{\pi d} = \frac{\mu_0(I_1+I_2)}{\pi d}$.

Two parallel conducting wires of equal length are placed distance $d$ apart and carry currents $I_1$ and $I_2$ respectively in opposite directions. The resultant magnetic field at the midpoint of the distance between both the wires is

Similar Questions

The magnetic field at the origin due to the current $I$ flowing in the wire is -

$A$ thin ring of radius $R$ metre has charge $q$ coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of $f$ revolution/s. The value of magnetic induction at the centre of the ring in $\text{Wb/m}^2$ is ($\mu_0$ = permeability of free space)

An electron revolves around a nucleus with rotational frequency $f$ in a circular orbit. Due to this,the magnetic induction produced at the nucleus position is $B$. The radius of the circular orbit is directly proportional to:

$AB$ and $CD$ are long straight conductors,separated by a distance $d$,each carrying a current $I$. The magnetic field at the midpoint of $BC$ is

$A$ current $I_0$ flows through a metallic circular loop of radius $r$. The resistance of the segment $ABC$ is half that of the segment $ADC$. The magnitude of the magnetic field at the centre $O$ of the loop is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module