
An 8N force acts on a rectangular conductor 20cm long placed perpendicular to a magnetic field. Determine the magnetic field induction if the current in the conductor is 40A.
Answer
133.8k+ views
Hint: Here a current carrying wire or conductor when placed in a magnetic field experiences a force. If the direction of the magnetic field and the direction of the wire or conductor are in 90 degrees with each other then the force that will be acting on the conductor would be perpendicular to both the magnetic field and the current carrying conductor. It can be determined using Fleming’s Left Hand Rule. Here apply the formula $F = BiL\sin \theta $; where F = Force; B = Magnetic Field Induction; i = current; L = Length of the conductor.
Complete step by step solution:
Put the known value in the formula and find B.
$F = BiL\sin \theta $
Take B to LHS and put rest of the variables in RHS
$\Rightarrow$ $\dfrac{F}{{iL\sin \theta }} = B$;
Put the given value in the above equation and solve,
$\Rightarrow$ $\dfrac{8}{{40 \times 0.2 \times \sin 90}} = B$; ….(Here $L = 20cm = 0.2m$)
Do the necessary mathematical calculation and solve for “B”.
$\Rightarrow$ $B = \dfrac{8}{8}$
The final value of B is:
$\Rightarrow$ $B = 1$ $Tesla$
The magnetic field induction is $B = 1$ $Tesla$.
Note: Here make sure to apply the correct formula for force on current carrying wire placed in a magnetic field. Do not use the formula $F = qvb\sin \theta $. Here we have been given the value of length of the wire and current in the wire. We need to use a formula ($F = BiL\sin \theta $) that relates all the given variables together.
Complete step by step solution:
Put the known value in the formula and find B.
$F = BiL\sin \theta $
Take B to LHS and put rest of the variables in RHS
$\Rightarrow$ $\dfrac{F}{{iL\sin \theta }} = B$;
Put the given value in the above equation and solve,
$\Rightarrow$ $\dfrac{8}{{40 \times 0.2 \times \sin 90}} = B$; ….(Here $L = 20cm = 0.2m$)
Do the necessary mathematical calculation and solve for “B”.
$\Rightarrow$ $B = \dfrac{8}{8}$
The final value of B is:
$\Rightarrow$ $B = 1$ $Tesla$
The magnetic field induction is $B = 1$ $Tesla$.
Note: Here make sure to apply the correct formula for force on current carrying wire placed in a magnetic field. Do not use the formula $F = qvb\sin \theta $. Here we have been given the value of length of the wire and current in the wire. We need to use a formula ($F = BiL\sin \theta $) that relates all the given variables together.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

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

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Wheatstone Bridge for JEE Main Physics 2025

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

Diffraction of Light - Young’s Single Slit Experiment

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

JEE Advanced 2024 Syllabus Weightage

JEE Main Chemistry Question Paper with Answer Keys and Solutions
