Question
Derive the expression for force per unit length between two long straight parallel current carrying conductors. Hence define one ampere.
✓
Answer
Figure shows two long parallel conductors $a$ and $b$ separated by $a$ distance $d$ and carrying $($parallel$)$ currents $I_a$ and $I _b,$ respectively.
The conductor $'a\ '$ produces, the same magnetic field $B$ a at all points along the conductor $'b\ '$.
The right-hand rule tells us that the direction of this field is downwards $($when the conductors are placed horizontally$)$. Its magnitude is given by
$\text{B}_{\alpha} = \frac{\mu_\circ\text{I}_{\alpha}}{2\pi\text{d}}$
The conductor $'b\ '$ carrying a current $I _b$ will experience a sideways force due to the field $B_a$.
The direction of this force is towards the conductor $'a\ '$.
We label this force as $F_{ba},$ the force on a segment $L$ of $'b\ '$ due to $'a\ '.$ The magnitude of this force is given by
$\int_{ba} = \text{I}_{b}\text{LB}_{a}$
$ = \frac{\mu_\circ\text{I}_{a}\text{I}_{b}}{2\pi\text{d}}\text{L}$
Let $\int_{ba}$ represent the magnitude of the force $\text{F}_{ba}$ per unit length.
$\int_{ba} = \frac{\mu_\circ\text{I}_{a}\text{I}_{b}}{2\pi\text{d}}$
One ampere: The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross $-$ section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to $2 \times 10^{–7}$ newton per metre of their length.
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The maximum energy of the emitted positron is $0.960 MeV$.
Given the mass values:
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calculate $Q$ and compare it with the maximum energy of the positron emitted.