Question

Question 25 (1 point) Two conductors are separated by 50 cm. The current flowing through one is 2A, and the current flowing through the other is 3 A. If the current is flowing in the same direction the magnetic filed between the conductors is 0 2.4x106T 1.6 x106T 0.8 x10-6 T 4x10-6 T

Answer 1

Given,

the separation between the conductors, d = 50cm = 50x10^–2 m

the current in the first conductor, I₁ = 2A

the current in the second conductor, I₂ = 3A

So, we have the formula for the magnetic field produced by a current carrying conductor = μ_oI/2πd

Since it is not specified at what distance between the conductors, the magnetic field is to be found, let us find the resultant magnetic field due to both the conductors, at midpoint (25cm) between the conductors.

Let the magnetic fields at the midpoint between the conductors, due to both the conductors, be B₁ and B₂ respectively.

It is given that the currents in the two conductors are in the same direction.

From Right Hand Thumb Rule, B₁ and B₂ will be in opposite directions at the midpoint.

Hence, the resultant field at the midpoint of two conductors, B = B₂ – B₁

= μ_oI₂/2π(d/2) – μ_oI₁/2π(d/2)

= μ_o/2π(d/2)[I₂ – I₁]   

= 2x10^–7/25x10^–2 [3 – 2] ( $\because$ μ_o/4π = 10^–7)

= 0.08x10^–5

B = 0.8x10^–6 T

So, the resultant field between the two conductors(at midpoint), is 0.8x10^–6 T.

Answer: 0.8x10^–6 T (Third option)