
The magnetic force on a current carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field is:
(A) $0$
(B) $ILB$
(C) $2ILB$
(D) $ILB/2$
Answer
140.1k+ views
Hint: The magnetic field due to a flowing charge or current carrying conductor is the product of the magnetic field at that point, the current flowing through the conductor and the length of the wire. It depends on the sine of the angle made by the magnetic field and the length of the wire.
Complete step by step solution:
In a magnetic field a current carrying wire experiences a magnetic force, which is given by –
$\vec F = I(d\vec l \times \vec B)$
Where I is the magnitude of current flowing in the conductor,
$dl$is the length of the wire
And B is the magnetic field.
Since the Force is also a vector quantity, the product between the length of wire and magnetic field is a cross product. Therefore, this formula can be rewritten as-
$\vec F = IBL\operatorname{Sin} \theta $
Where$\theta $is the angle between the magnetic field and the orientation of the wire in that field.
In the question we have,
$\theta = 90^\circ $as the wire is perpendicular to the magnetic field.
Therefore the force is given by,
$\vec F = IBL\operatorname{Sin} 90^\circ $
Since, $\sin 90^\circ = 1$
$F = IBL \times 1$
$F = IBL$
Thus, the answer is option (B).
Additional information
This formula for the force experienced by a conducting wire in a magnetic field is derived from the formula about moving charges experiencing in magnetic field as-
$\vec F = q\left( {\vec v \times \vec B} \right)$
Where q is the value of charge and v is its velocity.
Since the electric current is also a flow of charges, along a definite length, that is why it can be rewritten as-
$q\vec v = dq\dfrac{{dl}}{{dt}} = \dfrac{{dq}}{{dt}}dl = IL$
Note: The direction of the force is given by the right hand rule where if the index finger points In the direction if the current flowing inside the wire, and the middle finger points at the direction of the magnetic field line at the point, the direction of thumb gives the direction of the force experienced by the conductor.
Complete step by step solution:
In a magnetic field a current carrying wire experiences a magnetic force, which is given by –
$\vec F = I(d\vec l \times \vec B)$
Where I is the magnitude of current flowing in the conductor,
$dl$is the length of the wire
And B is the magnetic field.
Since the Force is also a vector quantity, the product between the length of wire and magnetic field is a cross product. Therefore, this formula can be rewritten as-
$\vec F = IBL\operatorname{Sin} \theta $
Where$\theta $is the angle between the magnetic field and the orientation of the wire in that field.
In the question we have,
$\theta = 90^\circ $as the wire is perpendicular to the magnetic field.
Therefore the force is given by,
$\vec F = IBL\operatorname{Sin} 90^\circ $
Since, $\sin 90^\circ = 1$
$F = IBL \times 1$
$F = IBL$
Thus, the answer is option (B).
Additional information
This formula for the force experienced by a conducting wire in a magnetic field is derived from the formula about moving charges experiencing in magnetic field as-
$\vec F = q\left( {\vec v \times \vec B} \right)$
Where q is the value of charge and v is its velocity.
Since the electric current is also a flow of charges, along a definite length, that is why it can be rewritten as-
$q\vec v = dq\dfrac{{dl}}{{dt}} = \dfrac{{dq}}{{dt}}dl = IL$
Note: The direction of the force is given by the right hand rule where if the index finger points In the direction if the current flowing inside the wire, and the middle finger points at the direction of the magnetic field line at the point, the direction of thumb gives the direction of the force experienced by the conductor.
Recently Updated Pages
JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address

JEE Main Course 2025 - Important Updates and Details

JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

A point charge + 20mu C is at a distance 6cm directly class 12 physics JEE_Main

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

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

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
