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
轉載自savemyexams
以上就是關于【CIE A Level Physics復習筆記20.1.4 Magnetic Flux Density】的解答,如需了解學校/賽事/課程動態,可至翰林教育官網獲取更多信息。
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