# 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 (b)(12aI_@)/17r (c)(aI_@)/r (d)(12aI_@)/35r

##### 1 Answer
Mar 14, 2018

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

#### Explanation:

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

${\mu}_{0} / \left\{4 \pi\right\} \frac{i}{d} \left(\sin {\phi}_{1} + \sin {\phi}_{2}\right)$

where ${\phi}_{1}$ and ${\phi}_{2}$ 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 $d$ of any arm of the "V" from the point O is gven by

$d \left(\cot {37}^{\circ} + \cot {63}^{\circ}\right) = a \implies d \left(\frac{4}{3} + \frac{3}{4}\right) = a \implies d = \frac{12}{25} a$.

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

${\mu}_{0} / \left\{4 \pi\right\} {I}_{2} / \left\{\frac{12 a}{25}\right\} \left(\sin {63}^{\circ} + \sin {37}^{\circ}\right) = {\mu}_{0} / \left\{4 \pi\right\} \frac{25 {I}_{2}}{12 a} \times \left(\frac{3}{5} + \frac{4}{5}\right) = {\mu}_{0} / \left\{\pi\right\} \frac{35}{48} {I}_{2} / a$

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 ${\mu}_{0} {I}_{1} / \left\{2 r\right\} = {\mu}_{0} / \pi {I}_{0} / \left\{2 r\right\}$ caused at O by the circular ring. Since the net field at O is zero, we have:

${\mu}_{0} / \left\{\pi\right\} \frac{35}{24} {I}_{2} / a = {\mu}_{0} / \pi {I}_{0} / \left\{2 r\right\} \implies {I}_{2} = \frac{12 a}{35 r} {I}_{0}$