The force acting per unit current per unit length on a current-carrying conductor placed perpendicular to the magnetic field

A straight conductor carrying a current of 1A normal to a magnetic field of flux density of 1 T with force per unit length of the conductor of 1 N m-1
A 15 cm length of wire is placed vertically and at right angels to a magnetic field.When a current of 3.0 A flows in the wire vertically upwards, a force of 0.04 N acts on it to the left.Determine the flux density of the field and its direction.
Step 1: Write out the known quantities
Force on wire, F = 0.04 N
Current, I = 3.0 A
Length of wire = 15 cm = 15 × 10-2 m
Step 2: Magnetic flux density B equation

Step 3: Substitute in values

Step 4: Determine the direction of the B field
Using Fleming’s left-hand rule :
F = to the left
I = vertically upwards
therefore, B = into the page
转载自savemyexams
以上就是关于【CIE A Level Physics复习笔记20.1.4 Magnetic Flux Density】的解答,如需了解学校/赛事/课程动态,可至翰林教育官网获取更多信息。
往期文章阅读推荐:
全网破防!ALevel CIE数学M1疑似错题?经济P2难度飙升?5月6日大考考情分析必看!
A-Level CIE就大规模泄题发布最严处罚!哪些考生必须重考?你的成绩怎么办?

© 2026. All Rights Reserved. 沪ICP备2023009024号-1