Question

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`

Answer 1

`I_2 = {12a}/{35r}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/{4pi} i/d (sin phi_1 +sin phi_2)`

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(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

`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 `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:

`mu_0/{pi} 35/24 I_2/a = mu_0/pi I_0/{2r} implies I_2 = {12a}/{35r}I_0`