A Helmholtz coil has a pair of loops,each wit | vedclass

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(B) The magnetic field on the axis of a circular coil with $N$ turns,radius $R$,and current $I$ at a distance $x$ from its center is given by $B = \fr

Vedclass · Accepted Answer

$\frac{8N{\mu _0}I}{5^{3/2}R}$

Vedclass · Answer

(B) The magnetic field on the axis of a circular coil with $N$ turns,radius $R$,and current $I$ at a distance $x$ from its center is given by $B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$.For a Helmholtz coil,the two loops are separated by a distance $R$. The point $P$ is at the midpoint,so the distance from each center ($A$ and $C$) to $P$ is $x = R/2$.The magnetic field at $P$ due to one loop is $B_1 = \frac{\mu_0 N I R^2}{2(R^2 + (R/2)^2)^{3/2}} = \frac{\mu_0 N I R^2}{2(R^2 + R^2/4)^{3/2}} = \frac{\mu_0 N I R^2}{2(5R^2/4)^{3/2}}$.Simplifying this,$B_1 = \frac{\mu_0 N I R^2}{2 \cdot (5/4)^{3/2} \cdot R^3} = \frac{\mu_0 N I}{2 \cdot (5^{3/2}/8) \cdot R} = \frac{4 \mu_0 N I}{5^{3/2} R}$.Since the currents flow in the same direction,the magnetic fields from both loops at $P$ are in the same direction. Therefore,the total magnetic field is $B = 2 B_1 = 2 \cdot \frac{4 \mu_0 N I}{5^{3/2} R} = \frac{8 \mu_0 N I}{5^{3/2} R}$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{r} \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$
$(B). \frac{\mu_0 i}{2r} \odot$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$
$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$
$(D). \frac{\mu_0 i}{4r} \odot$	$(D). 0$	$(D). 0$

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