Class 12 Physics
CBSE and UP Board
Moving Charges and Magnetism
Chapter-4 Exercise 4.3
NCERT Solutions Class 12 Physics Chapter 4 Question 3
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
Therefore magnitude of the distance of the point from the wire, r = 2.5 m.
Magnitude of the magnetic field at that point is given by the relation,
|B|= (μ0/4π) 2l/r
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
|B|= (4π x 10⁻⁷ )/4π x (2x 50 )/2.5 = 4 x 10⁻⁶ T
The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.