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(D) The magnetic field $B$ due to a finite straight wire of length $L$ at a perpendicular distance $d$ from its centre is given by $B = \frac{\mu_0 i}

Vedclass · Accepted Answer

$8$

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(D) The magnetic field $B$ due to a finite straight wire of length $L$ at a perpendicular distance $d$ from its centre is given by $B = \frac{\mu_0 i}{4 \pi d} (\sin 	heta_1 + \sin 	heta_2)$.For a square loop of side $a = \frac{1}{\sqrt{2}} \; m$,the perpendicular distance $d$ from the centre to any side is $d = \frac{a}{2} = \frac{1}{2\sqrt{2}} \; m$.The angles subtended by the ends of the side at the centre are $	heta_1 = 45^\circ$ and $	heta_2 = 45^\circ$.Substituting the values: $B_{	ext{side}} = \frac{4 \pi 	imes 10^{-7} 	imes 5}{4 \pi 	imes (\frac{1}{2\sqrt{2}})} (\sin 45^\circ + \sin 45^\circ)$.$B_{	ext{side}} = \frac{10^{-7} 	imes 5}{\frac{1}{2\sqrt{2}}} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = 10^{-7} 	imes 5 	imes 2\sqrt{2} 	imes \frac{2}{\sqrt{2}} = 20 	imes 10^{-7} = 2 	imes 10^{-6} \; T$.The total magnetic field at the centre due to all $4$ sides is $B_{	ext{net}} = 4 	imes B_{	ext{side}} = 4 	imes 2 	imes 10^{-6} = 8 	imes 10^{-6} \; T$.Comparing this with $p 	imes 10^{-6} \; T$,we get $p = 8$.

$A$ current of $5 \; A$ flows through a square loop of side length $\frac{1}{\sqrt{2}} \; m$. The magnitude of the magnetic field $B$ at the centre of the square loop is $p \times 10^{-6} \; T$. Find the value of $p$. [Take $\mu_0 = 4 \pi \times 10^{-7} \; T \cdot m \cdot A^{-1}$].

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