Question

In Fig. 29-65, a long straight wire carries a current i₁ = 27.8 A and a rectangular loop carries current i₂ = 19.6 A. Take a = 1.05 cm, b = 7.13 cm, and L = 22.8cm. What is the magnitude of the net force on the loop due to i₁?



Number__________Units N

Answer 1

There is no force will act on the left and right sides of the

rectangular loop due to the magnetic field produced by

the current \(I_{1}\).

Magnetic force acting on the upper portion of the loop due to

the current carrying straight wire is

$$ \begin{aligned} F_{1} &=\frac{\mu_{0}^{i} 1^{i} 2^{L}}{2 \pi a} \\ &=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(27.8 \mathrm{~A})(19.6 \mathrm{~A})(0.228 \mathrm{~m})}{2 \pi\left(1.05 \times 10^{-2} \mathrm{~m}\right)} \\ &=2.37 \times 10^{-3} \mathrm{~N} \end{aligned} $$

Magnetic force acting on the lower portion of the loop due to

the current carrying straight wire is

$$ \begin{aligned} F_{2} &=\frac{\mu_{0} i_{1} i_{2} L}{2 \pi(a+b)} \\ &=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(27.8 \mathrm{~A})(19.6 \mathrm{~A})(0.228 \mathrm{~m})}{2 \pi\left(1.05 \times 10^{-2} \mathrm{~m}+7.13 \times 10^{-2} \mathrm{~m}\right)} \\ &=0.304 \times 10^{-3} \mathrm{~N} \end{aligned} $$

These forces, \(F_{1}\) and \(F_{2}\) are in the opposite directions, hence net

magnetic force acting on the loop is

$$ \begin{aligned} F &=2.37 \times 10^{-3} \mathrm{~N}-0.304 \times 10^{-3} \mathrm{~N} \\ &=2.07 \mathrm{mN} \end{aligned} $$