Question 262: Magnetism and Matter

Question

A current of 5 A exists in a square loop of side $\frac{1}{\sqrt{2}} \mathrm{~m}$. Then the magnitude of the magnetic field $B$ at the centre of the square loop will be $p 	imes 10^{-6} \mathrm{~T}$. where, value of p is ________ $\left[ight.$ Take $\mu_0=4 \pi 	imes 10^{-7} \mathrm{~T} \mathrm{~mA}^{-1}$ ].

Accepted Answer

$$ B = \frac{2 \sqrt{2}\,\mu_0 I}{\pi L} $$
For a square loop of side length
$$ L = \frac{1}{\sqrt{2}} 	ext{ m}, $$
the perpendicular distance from the center to any side is
$$ d = \frac{L}{2} = \frac{1}{2\sqrt{2}} 	ext{ m}. $$
For a finite straight wire, the magnetic field at a point at distance $ d $ from the wire is given by the BiotÃ¢ÂÂSavart expression
$$ B_{	ext{side}} = \frac{\mu_0 I}{4\pi d} \left(\sin	heta_1 + \sin	heta_2ight) $$
where $ 	heta_1 $ and $ 	heta_2 $ are the angles between the extended wire and the line joining the ends of the wire with the observation point. For one side of the square, by symmetry, the midpoint of the side and the center of the square give
$$ 	an	heta = \frac{L/2}{L/2} = 1 \quad\Rightarrow\quad 	heta = 45^\circ. $$
Thus,
$$ \sin	heta_1 = \sin	heta_2 = \sin 45^\circ = \frac{1}{\sqrt{2}}. $$
Then the field due to one side is
$$ B_{	ext{side}} = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}ight)} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}ight) = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}ight)} \left(\frac{2}{\sqrt{2}}ight) = \frac{\mu_0 I \sqrt{2}}{2\pi L}. $$
Since there are 4 sides with contributions that add vectorially in the same direction at the center, the total magnetic field is
$$ B = 4 \cdot B_{	ext{side}} = \frac{4 \mu_0 I \sqrt{2}}{2\pi L} = \frac{2 \sqrt{2}\,\mu_0 I}{\pi L}. $$
Substitute
$$ L = \frac{1}{\sqrt{2}}, $$
to obtain
$$ B = \frac{2 \sqrt{2}\,\mu_0 I}{\pi \left(\frac{1}{\sqrt{2}}ight)} = \frac{2 \sqrt{2}\,\mu_0 I \sqrt{2}}{\pi} = \frac{4\,\mu_0 I}{\pi}. $$
Using the provided values
$$ I = 5 	ext{ A} \quad 	ext{and} \quad \mu_0 = 4\pi 	imes 10^{-7} 	ext{ T\,m/A}, $$
the magnetic field becomes
$$ B = \frac{4 	imes \left(4\pi 	imes 10^{-7}ight) 	imes 5}{\pi} = \frac{80\pi 	imes 10^{-7}}{\pi} = 80 	imes 10^{-7} 	ext{ T} = 8 	imes 10^{-6} 	ext{ T}. $$
Thus, the value of $ p $ is
$$ p = 8. $$

Magnetism and Matter

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