A circular loop wire with current I1=I∘π and a V shaped wire with current I2 are arranged in a plane as shown in diagram. If magnetic field at O is zero, value of I2 is ? (a) $\frac{13 a I_{\circ}}{35} r$ $(13aI_@)/35r$ (b) $\frac{12 a I_{\circ}}{17} r$ $(12aI_@)/17r$ (c) $\frac{a I_{\circ}}{r}$ $(aI_@)/r$ (d) $\frac{12 a I_{\circ}}{35} r$ $(12aI_@)/35r$

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A circular loop wire with current I1=I∘π and a V shaped wire with current I2 are arranged in a plane as shown in diagram. If magnetic field at O is zero, value of I2 is ? (a) $\frac{13 a I_{\circ}}{35} r$ $(13aI_@)/35r$ (b) $\frac{12 a I_{\circ}}{17} r$ $(12aI_@)/17r$ (c) $\frac{a I_{\circ}}{r}$ $(aI_@)/r$ (d) $\frac{12 a I_{\circ}}{35} r$ $(12aI_@)/35r$

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Answer 1 · 2018-03-14T00:59:12

$I_{2} = \frac{12 a}{35 r} I_{0}$ $I_2 = {12a}/{35r}I_0$ (I have assumed that the question reads $I_{1} = \frac{I_{0}}{π}$ $I_1=I_0/pi$ )

Explanation:

The magnetic field due to a straight line segment carrying a current $i$ $i$ , at a point at a distance $d$ $d$ from the wire is given by

$\frac{μ_{0}}{4 π} \frac{i}{d} ({sin ϕ}_{1} + {sin ϕ}_{2})$ $mu_0/{4pi} i/d (sin phi_1 +sin phi_2)$

where $ϕ_{1}$ $phi_1$ and $ϕ_{2}$ $phi_2$ are the angles that the lines joining the ends of the wire to the point makes with the perpendicular dropped to the wire from the point.

The distance $d$ $d$ of any arm of the "V" from the point O is gven by

$d ({cot 37}^{\circ} + {cot 63}^{\circ}) = a \Rightarrow d (\frac{4}{3} + \frac{3}{4}) = a \Rightarrow d = \frac{12}{25} a$ $d(cot37^circ+cot63^circ) = a implies d(4/3+3/4)=a implies d=12/25a$ .

So, the magnetic field due to one arm of the "V" at the point O is given by

$\frac{μ_{0}}{4 π} \frac{I_{2}}{\frac{12 a}{25}} ({sin 63}^{\circ} + {sin 37}^{\circ}) = \frac{μ_{0}}{4 π} \frac{25 I_{2}}{12 a} \times (\frac{3}{5} + \frac{4}{5}) = \frac{μ_{0}}{π} \frac{35}{48} \frac{I_{2}}{a}$ $mu_0/{4pi} I_2/{(12a)/25} (sin 63^circ +sin 37^circ) = mu_0/{4pi} {25I_2}/{12a} times (3/5+4/5) = mu_0/{pi} 35/48 I_2/a$

The field due to the "V" is twice of this (each arm contributing the same amount), and this points downwards - as opposed to the upwards field $μ_{0} \frac{I_{1}}{2 r} = \frac{μ_{0}}{π} \frac{I_{0}}{2 r}$ $mu_0 I_1/{2r} = mu_0/pi I_0/{2r}$ caused at O by the circular ring. Since the net field at O is zero, we have:

$\frac{μ_{0}}{π} \frac{35}{24} \frac{I_{2}}{a} = \frac{μ_{0}}{π} \frac{I_{0}}{2 r} \Rightarrow I_{2} = \frac{12 a}{35 r} I_{0}$ $mu_0/{pi} 35/24 I_2/a = mu_0/pi I_0/{2r} implies I_2 = {12a}/{35r}I_0$

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A circular loop wire with current I1=I∘π and a V shaped wire with current I2 are arranged in a plane as shown in diagram. If magnetic field at O is zero, value of I2 is ? (a)13aI∘35r(13aI_@)/35r (b)12aI∘17r(12aI_@)/17r (c)aI∘r(aI_@)/r (d)12aI∘35r(12aI_@)/35r

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