Question

Figure shows a part of an electric circuit. The wires AB, CD and EF are long and have identical resistance. The separation between the neighbouring wires is 1.0 cm. The wires AE and BF have negligible resistance and the ammeter reads 30 A. Calculate the magnetic force per unit length of AB and CD.
Figure

Answer 1

Explanation:

To find: The magnetic force per unit length of AB and CD.

Because of wires AB, CD and EF have the same resistance, the current (30 A) is distributed equally in them, that is to say 10 A in each cable.  

Due to a parallel current-bearing wire, the magnetic force per unit length is given by

$\frac{F}{l}=\frac{\mu_{o} i_{1} i_{2}}{2 \pi d}$

Magnetic force per unit length AB = Force due to current CD + Force compared to current in EF

$\frac{F}{l}=\frac{\mu_{0} \times 10 \times 10}{2 \pi \times 1 \times 10^{-2}}+\frac{\mu_{0} \times 10 \times 10}{2 \pi \times 2 \times 10^{-2}}$

$=\frac{\mu_{0} \times 100}{2 \pi \times 10^{-2}}+\frac{\mu_{0} \times 100}{2 \pi \times 2 \times 10^{-2}}$

$=\frac{100 \mu_{0}}{2 \pi \times 10^{-2}}+\frac{100 \mu_{0}}{2 \pi \times 2 \times 10^{-2}}$

$=\frac{2 \times 10^{-7} \times 100}{10^{-2}}+\frac{2 \times 10^{-7} \times 100}{2 \times 10^{-2}}$

$=3 \times 10^{-3} N / M$

Similarly,

Magnetic force per unit length of CD = Force from current in AB − Force from current in EF

CD force due to current in AB = Current force in EF  

Magnetic force per unit length of CD = 0

Answer 2

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