Question

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(a) A solenoid is of length L and has n turns. i. Write the expression for the magnetic flux density B within the solenoid 11. State, in terms of B, the magnetic flux density at the ends of the solenoid when the current in the solenoid is 1 [2 marks] (b) Two long straight wires are separated by a distance R. Both wires carry the same current / but in opposite direction. Point P is located midway between the two wires. What is the magnetic field strength at P? 5 marks] Two long parallel straight wires X and Y are placed d m apart in a vacuum. The current flowing in wires X and Y are I and 3/respectively and the force per unit length on each wire is kNm1. The current flowing in the wires X and Y are then increased to 2/ and 4/ respectively. What is the distance between the wires so that the force per unit length is 3kN m1? (c) 3 marks

Answer 1

(a) A solenoid is of length 'L' and has 'n' turns.

i. Write an expression for the magnetic flux density 'B' within the solenoid when the current in the solenoid is 'I₁'.

$\Rightarrow$ The magnetic field within the solenoid has a uniform value 'B' and outside the coils has a value 'zero'.

Using an ampere's law, we have

_$\oint$$\vec{B}$ . d$\vec{s}$ = _$\mu$0 I_enc

where, I_enc = current enclosed = I₁ (n L)

B L = _$\mu$0 I₁ n L

B = _$\mu$0 n I₁

(b) Two long straight wires are separated by a distance R. Both wires carry the same current 'I' but in opposite direction.

Wire 1 Wire 2 (R/2) (R/2

The magnetic field strength at point 'P' which will be given as -

B_P = B₁ + B₂

B_P = (_$\mu$0 I₁ / d₁) + (_$\mu$0 I₂ / d₂)

where, d₁ = d₂ = R/2

I₁ = I₂ = I

then, we get

B_P = [_$\mu$0 I / (R/2)] - [_$\mu$0 I / (R/2)]

B_P = 0

(c) Force between two wires which will be given by -

F / L = _$\mu$0 I₁ I₂ / 2$\pi$ d

(k N/m) = (4$\pi$ x 10^-7 mT/A) (I) (3 I) / 2$\pi$ d

(k N/m) = (6 x 10^-7 mT) I² / d

d = (6 x 10^-7 mT) I² / (k N/m)

What is the distance between the wires so that the force per unit length is 3k N/m?

using a formula, we have

F / L = _$\mu$0 I₁ I₂ / 2$\pi$ d

(3k N/m) = (4$\pi$ x 10^-7 mT/A) (2I) (4 I) / 2$\pi$ d

(3k N/m) = (16 x 10^-7 mT) I² / d

d = (16 x 10^-7 mT) I² / (3k N/m)