## Chapter 7

## Q. 7.2

## Step-by-Step

## Verified Solution

At point P

H_{1} (I_{1}) = \frac{50}{2π (15 + 10) × 10^{-2} } A/m

H_{2} (I_{2}) = \frac{I_{2}}{2π × 10 × 10^{-2} } A/m

For net H at P to be zero

H_{1} + H_{2} = 0 or H_{2} = -H_{1}

or \frac{I_{2}}{2π × 10 × 10^{-2} } = \frac{50}{2π × 10 × 10^{-2}}

or

I_{2} = – 50 × \frac{10}{25} = –20 A (in opposite direction to I_{1})