
The figure shows a wire sliding on two parallel conducting rails placed at a separation $\prime I\prime $ . A magnetic field $B$ exists in a direction perpendicular to the plane of the rails. What force is necessary to keep the wire moving at a constant velocity $v$ ?
(A) $\dfrac{{{B^2}{l^2}v}}{R}$
(B) $\dfrac{{2{B^2}{l^2}v}}{R}$
(C) $\dfrac{{{B^2}{l^2}v}}{{2R}}$
(D) None of these

Answer
125.4k+ views
Hint: To solve this question, we will use the expression for magnetic Lorentz force. Also, we will need ohm’s law and Faraday's law for induced emf to find the current induced in the circuit as there is a change in the magnetic field, which results in generating induced emf in a closed loop. Then, by substituting the expression for the current in the Lorentz force equation, we can find the force.
Formula used:
$F = Bil$
$\varepsilon = Blv$
$i = \dfrac{\varepsilon }{R}$
Complete step-by-step solution:
The force experienced by the wire while moving on the parallel conducting wire through the magnetic field is given by the equation,
$F = i\left( {\vec B \times \vec l} \right)$
$F = Bil$
where $i$ is the induced current flowing through the loop.
$\vec B$ is the magnetic field.
and $\vec l$ is the length of the wire.
Now, the current developed inside the loop can be evaluated using ohm’s law.
$i = \dfrac{\varepsilon }{R}$
where emf say $\varepsilon $ is the induced emf due to change in magnetic flux as the wire is moving
and $R$ is the resistance of the loop.
Now, the induced emf is given by faraday’s law for metallic wire as,
$\varepsilon = \dfrac{{d\phi }}{{dt}}$
$ \Rightarrow \varepsilon = \dfrac{{d\left( {Blx} \right)}}{{dt}}$
where $x$ is the distance moved by the wire
$ \Rightarrow \varepsilon = Bl\dfrac{{dx}}{{dt}}$
$\therefore \varepsilon = Blv$
where, $v$ is the velocity of the wire.
At this time, substituting this value for emf in ohm’s law, we will get the current induced in the loop.
$i = \dfrac{\varepsilon }{R}$
$ \Rightarrow i = \dfrac{{Blv}}{R}$
Currently, we can compute the force essential to keep the wire moving with a constant velocity by substituting this current in the expression for Lorentz force.
$F = Bil$
Put the value of $i$ in the above equation we get,
$F = B\left( {\dfrac{{Blv}}{R}} \right)l$
$ \Rightarrow F = \dfrac{{{B^2}{l^2}v}}{R}$
Hence, the force required is found to be $\dfrac{{{B^2}{l^2}v}}{R}$ .
So, option (A) is correct.
Note: We must be having a clear memory of formulas and theories of basic laws and equations to solve. Moreover, these types of questions may come as an open circuit as an alternative to closed-loop as this question. Then no current will be flowing over the circuit and force becomes zero.
Formula used:
$F = Bil$
$\varepsilon = Blv$
$i = \dfrac{\varepsilon }{R}$
Complete step-by-step solution:
The force experienced by the wire while moving on the parallel conducting wire through the magnetic field is given by the equation,
$F = i\left( {\vec B \times \vec l} \right)$
$F = Bil$
where $i$ is the induced current flowing through the loop.
$\vec B$ is the magnetic field.
and $\vec l$ is the length of the wire.
Now, the current developed inside the loop can be evaluated using ohm’s law.
$i = \dfrac{\varepsilon }{R}$
where emf say $\varepsilon $ is the induced emf due to change in magnetic flux as the wire is moving
and $R$ is the resistance of the loop.
Now, the induced emf is given by faraday’s law for metallic wire as,
$\varepsilon = \dfrac{{d\phi }}{{dt}}$
$ \Rightarrow \varepsilon = \dfrac{{d\left( {Blx} \right)}}{{dt}}$
where $x$ is the distance moved by the wire
$ \Rightarrow \varepsilon = Bl\dfrac{{dx}}{{dt}}$
$\therefore \varepsilon = Blv$
where, $v$ is the velocity of the wire.
At this time, substituting this value for emf in ohm’s law, we will get the current induced in the loop.
$i = \dfrac{\varepsilon }{R}$
$ \Rightarrow i = \dfrac{{Blv}}{R}$
Currently, we can compute the force essential to keep the wire moving with a constant velocity by substituting this current in the expression for Lorentz force.
$F = Bil$
Put the value of $i$ in the above equation we get,
$F = B\left( {\dfrac{{Blv}}{R}} \right)l$
$ \Rightarrow F = \dfrac{{{B^2}{l^2}v}}{R}$
Hence, the force required is found to be $\dfrac{{{B^2}{l^2}v}}{R}$ .
So, option (A) is correct.
Note: We must be having a clear memory of formulas and theories of basic laws and equations to solve. Moreover, these types of questions may come as an open circuit as an alternative to closed-loop as this question. Then no current will be flowing over the circuit and force becomes zero.
Recently Updated Pages
JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

The formula of the kinetic mass of a photon is Where class 12 physics JEE_Main

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Main Login 2045: Step-by-Step Instructions and Details

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Ideal and Non-Ideal Solutions Raoult's Law - JEE

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
