A long straight wire in the horizontal plane carries a current of 50 Ain north to south direction. Give the magnitude and direction of Bat a point 2.5 m east of the wire.
Given: The current in the wire is 50 A , a point is 2.5 m away from the east of the wire, the distance of a point from the wire is 2.5 m .
The magnitude of the magnetic field at this point is given as,
| B → | = μ 0 4 π 2 I r
Where, the current in the wire is I , the distance of the point from the wire is r and the permeability of free space is μ 0 .
By substituting the given values in the above expression, we get
| B → | = 4 π × 10 − 7 4 π 2 × 50 2.5 = 4 × 10 − 6 T
Since, the direction of current in the wire is north to south (downward), then from the right hand thumb rule, the direction of magnetic field east of the wire is vertically upward direction.
Given: The current in the wire is
The magnitude of the magnetic field at this point is given as,
Where, the current in the wire is
By substituting the given values in the above expression, we get
Since, the direction of current in the wire is north to south (downward), then from the right hand thumb rule, the direction of magnetic field east of the wire is vertically upward direction.
