A straight horizontal copper rod carries a current of 50.0 A from west to east in a region between the poles of a large electromagnet. In this region there is a horizontal magnetic field toward the northeast (that is,45 north of east) with magnitude 1.20T. Find the magnitude and direction of the force on a 1.00 m section of rod
Answers
Answer:
A straight horizontal copper rod carries a current of 50.0 A from west to east in a region between the poles of a large electromagnet. In this region there is a horizontal magnetic field toward the north-east (that is 45° north of east) with magnitude 1.2 T.
Given:
current I = 50.0 A
magnetic field B = 1.20T
lengh of rod L = 1.00 m
angle θ = 45°
To find:
magnitude and direction of the force F =?
Step-to-step-explanation:
- When a copper rod carries a current is placed between magnetic field then the force is produced on the rod. This force is given by:
F = ILB sinθ
where
F = force acting on the rod
I = current in the rod
L = lengh of rod
B = magnetic field in which rod is placed
θ = the angle between current & magnetic field
- Put the given values in above formula
F = 50 × 1 × 1.20 × sin 45°
F = 42.4 N
- The direction of the force is perpendicular to the plane of current & the magnetic field.
- The plane of current & the field are lie on horizontal plane then the direction of force will be upword.
- According of right hand thumb rule, the direction of force is vertically upword.
Hence the the magnitude of force is 42.4 N & the direction of force is vertically upword.