Test: Ampere Circuital Law(3 Dec) - JEE MCQ

The line integral of the magnetic field of induction  around any closed path in free space is equal to absolute permeability of free space μ₀ times the total current flowing through area bounded by the path.
Ampere's circuital law is given by: 

From Ampere circuital law

The direction of field at the given point will be vertical up determined by the screw rule or right hand rule.

I = 40A
r = 15 cm = 15 x 10^-2 m
∴  = 5.34 x 10^-5 T

The magnetic field from the centre of wire of radius R is given by
B = ((μ₀I)/(2R²))r (r < R) ⇒ B ∝ r
and B = μ₀I/2πr (r > R) ⇒ B ∝ 1/r
From this descriptions, we can say that the graph (b) is a correct representation.

Here, n = 600/0.6 = 1000 turns/m, I = 4A
l = 0.6m, r = 0.02 m ∵ 1/r = 30 i.e., l >>r
Hence, we can use long solenoid formuls, then
∴ B = μ₀nI = 4π x 10^-7 x 10³ x 4 = 50.24 x 10^-4 
= 5.024 x 10^-3 T

Here, I = 2.5 A , I = 50 cm = 0.50 m and n = 100/0.50 = 200 m^-1
∴ 

For six layers of windings the total number of turns = 6 x 450 = 2700
Now number of turns per unit length 

Then the field in side the solenoid near the centre 
B = μ₀nI = 4π x 10^-7 x 3000 x 6  = 72π x 10^-4 T = 72πG

Option C is incorrect as magnetic field inside the core of the toroid varies greatly.

Consider two amperian loops of radius a/ 2 and 2 a as shown in the diagram. Applying ampere's circuital law for these loops, we get

For the smaller loop,

Now
Here OP
Now, and