(C) The force per unit length between two parallel long conductors is given by the formula: $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.Initially,the force is

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

(C) The force per unit length between two parallel long conductors is given by the formula: $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.Initially,the force is $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.When the current in one conductor is doubled $(I_1' = 2I_1)$ and its direction is reversed,the new current becomes $-2I_1$. The distance is increased to $d' = 3d$.The new force $F'$ is given by: $F' = \frac{\mu_0 (-2I_1) I_2}{2 \pi (3d)}$.Simplifying this,we get: $F' = -\frac{2}{3} \left( \frac{\mu_0 I_1 I_2}{2 \pi d} \right)$.Substituting the initial force $F$,we get: $F' = -\frac{2F}{3}$.

Class 12 Physics Moving Charges and Magnetism Force on a Current Carrying Conductor

Medium

Two long conductors separated by a distance $d$ carry currents $I_1$ and $I_2$ in the same direction. They exert a force $F$ on each other. Now,the current in one of them is increased to $2$ times and its direction is reversed. The distance between them is also increased to $3d$. The new value of force between them is:

MHT CET 2023 All MHT CET

MHT CET 2021 All MHT CET

A
$-2F$
B
$\frac{F}{3}$
C
$\frac{-2F}{3}$
D
$\frac{-F}{3}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Moving Charges and Magnetism

Subtopic

Force on a Current Carrying Conductor

Previous Year

MHT CET 2023 Paper All MHT CET years →

Previous Year

MHT CET 2021 Paper All MHT CET years →

$A$ square loop,carrying a steady current $I$,is placed in a horizontal plane near a long straight conductor carrying a steady current $I_1$ at a distance $d$ from the conductor as shown in the figure. The loop will experience:

MediumAIPMT 2011

View Solution

$A$ straight wire of length $0.5\,m$ carrying a current of $1.2\,A$ is placed in a uniform magnetic field of induction $2\,T$. The magnetic field is perpendicular to the length of the wire. The force on the wire is.......$N$.

EasyAIPMT 1992

View Solution

Assertion $(A):$ $A$ wire is bent into an irregular shape with the points $P$ and $Q$ fixed. If a current $I$ is passed through the wire,then the area enclosed by the irregular portion of the wire increases.
Reason $(R):$ Wires carrying currents in opposite directions repel each other.

MediumAIIMS 2015

View Solution

$A$ current of $5 \ A$ is flowing in a wire of length $1.5 \ m$. $A$ force of $7.5 \ N$ acts on it when it is placed in a uniform magnetic field of $2 \ T$. The angle between the magnetic field and the direction of the current is......$^o$

Easy

View Solution

$A$ straight current-carrying conductor is placed such that the current flows out of the plane of the paper. The conductor is placed between the poles of two magnets,as shown in the figure. The conductor will experience a force in the direction towards:

Difficult

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$A$ square loop,carrying a steady current $I$,is placed in a horizontal plane near a long straight conductor carrying a steady current $I_1$ at a distance $d$ from the conductor as shown in the figure. The loop will experience:

$A$ straight wire of length $0.5\,m$ carrying a current of $1.2\,A$ is placed in a uniform magnetic field of induction $2\,T$. The magnetic field is perpendicular to the length of the wire. The force on the wire is.......$N$.

Assertion $(A):$ $A$ wire is bent into an irregular shape with the points $P$ and $Q$ fixed. If a current $I$ is passed through the wire,then the area enclosed by the irregular portion of the wire increases.Reason $(R):$ Wires carrying currents in opposite directions repel each other.

$A$ current of $5 \ A$ is flowing in a wire of length $1.5 \ m$. $A$ force of $7.5 \ N$ acts on it when it is placed in a uniform magnetic field of $2 \ T$. The angle between the magnetic field and the direction of the current is......$^o$

$A$ straight current-carrying conductor is placed such that the current flows out of the plane of the paper. The conductor is placed between the poles of two magnets,as shown in the figure. The conductor will experience a force in the direction towards:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion $(A):$ $A$ wire is bent into an irregular shape with the points $P$ and $Q$ fixed. If a current $I$ is passed through the wire,then the area enclosed by the irregular portion of the wire increases.
Reason $(R):$ Wires carrying currents in opposite directions repel each other.