Test: Magnetostatics - 1 - Electronics and Communication Engineering (ECE) MCQ

Given, r = 1 m, H = 1 A/m
Using Ampere’s circuit law, we have:

The magnetic flux density at the centre of the square having each side a is given by

Given, a = 2 m, I = 5 Amp 

Given, l = 20 cm = 0.2 m,
r - 0.5 cm = radius = 0.005 m
N = 1000,
B = 2 Wb/m²,
I = 10 A
The torque acting on the solenoid is
T = NIAB sinθ
T will be maximum when
θ = 90°, i.e. sin90° = 1
∴ T_max = 1000 x 10 (0.005 x 0.2) x 2
= 20 N-m

Magnetic flux density at the centre of a current carrying loop is given by

Curl  cartesian coordinate is given by


Since H_x = constant and H_y = H_z = 0, therefore


= 0 + 0 + 0 

Let medium-1 and medium-2 are respectively iron and air, then

or  
or  

• Both assertion and reason are correct statements.
• The potential in case of magnetic field is a vector potential because source for producing a magnetic field is a current element which has both magnitude and direction.
• The potential in case of electric field i.e. electric potential is a scalar potential because the source for producing an electric field is a charge which is a scaler quantity (having only magnitude).