Two infinitely long parallel wires carry curr | vedclass

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

Question

(A) Let the magnetic field be minimum at point $P$,at a distance $x$ from the first wire.The magnetic field due to the first wire at $P$ is $B_1 = \fr

Vedclass · Accepted Answer

$\frac{I_2}{I_1}=9$,anti-parallel

Vedclass · Answer

(A) Let the magnetic field be minimum at point $P$,at a distance $x$ from the first wire.The magnetic field due to the first wire at $P$ is $B_1 = \frac{\mu_0 I_1}{2 \pi x}$.The magnetic field due to the second wire at $P$ is $B_2 = \frac{\mu_0 I_2}{2 \pi (d-x)}$,where $d = 4 \, cm$.For the net magnetic field to be a non-zero minimum between the wires,the fields must be in opposite directions,which implies the currents must be anti-parallel.The net magnetic field is $B = |B_1 - B_2| = \frac{\mu_0}{2 \pi} |\frac{I_1}{x} - \frac{I_2}{d-x}|$.For $B$ to be minimum,the derivative $\frac{dB}{dx} = 0$.$\frac{d}{dx} (\frac{I_1}{x} - \frac{I_2}{d-x}) = 0 \Rightarrow -\frac{I_1}{x^2} - \frac{I_2}{(d-x)^2} = 0$.This implies $\frac{I_1}{x^2} = -\frac{I_2}{(d-x)^2}$. Since $I_1, I_2 > 0$,this confirms the currents are anti-parallel.Taking magnitudes: $\frac{I_2}{I_1} = \frac{(d-x)^2}{x^2}$.Given $d = 4 \, cm$ and $x = 1 \, cm$,we have $\frac{I_2}{I_1} = \frac{(4-1)^2}{1^2} = \frac{3^2}{1^2} = 9$.Thus,the ratio $\frac{I_2}{I_1} = 9$ and the currents are anti-parallel.

Similar Questions

Two circular loops having same radius $[ R = 10 \, cm ]$ and same current $[ I = \frac{7}{2} \, A ]$ are placed along the same axis as shown. If the distance between their centers is $[ 10 \, cm ]$,find the net magnetic field at point $P$ located at the midpoint between them.

Current through $ABC$ and $A^{\prime} B^{\prime} C^{\prime}$ is $I$. What is the magnetic field at $P$? $BP = PB^{\prime} = r$. (Here $C^{\prime} B^{\prime} PBC$ are collinear).

$A$ straight wire of length $20 \text{ cm}$ carrying a current of $\frac{3}{\pi^2} \text{ A}$ is bent in the form of a circle. The magnetic field at the centre of the circle is

$A$ 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 $2a$ from the axis of the wire is:

$A$ steady current flows in a long wire. It is bent into a circular loop of one turn and the magnetic field at the centre of the coil is $B$. If the same wire is bent into a circular loop of $n$ turns,the magnetic field at the centre of the coil is

Vedclass Test Series

Exam Paper Generator

Online Exam Module