Question

A circular loop of radius 10 cm carrying current of 1.0 A lies in the x-y plane. A long straight wire lies in the same plane parallel to the x-axis at a distance of 20 cm as shown in the figure. 

Find the direction and value of current that has to be maintained in the wire so that the net magnetic field at O is zero. 
 

Hint:

For net zero magnetic field at a point, the contributions from different sources should cancel each other.

Answer 1

Applying Biot-Savart Law and Ampere’s Law 
- The magnetic field at the center \( O \) of the circular loop is given by: \[ B_{\text{loop}} = \frac{\mu_0 I_{\text{loop}}}{2R} \] where \( I_{\text{loop}} = 1.0 A \), \( R = 10 \) cm = 0.1 m. 
\[ B_{\text{loop}} = \frac{(4\pi \times 10^{-7}) (1)}{2 \times 0.1} \] \[ B_{\text{loop}} = 2 \times 10^{-6} \text{ T} \] - The magnetic field due to the long straight wire at distance \( d = 20 \) cm is: \[ B_{\text{wire}} = \frac{\mu_0 I_{\text{wire}}}{2\pi d} \] \[ B_{\text{wire}} = \frac{(4\pi \times 10^{-7}) I_{\text{wire}}}{2\pi \times 0.2} \] \[ B_{\text{wire}} = \frac{2 \times 10^{-7} I_{\text{wire}}}{0.2} \] \[ B_{\text{wire}} = 10^{-6} I_{\text{wire}} \] - For net magnetic field at \( O \) to be zero: \[ B_{\text{loop}} = B_{\text{wire}} \] \[ 2 \times 10^{-6} = 10^{-6} I_{\text{wire}} \] \[ I_{\text{wire}} = 2 A \] Direction of Current in the Wire: 
- Using the Right-Hand Rule, the circular loop produces a magnetic field into the plane at O. 
- To cancel it, the wire must produce a magnetic field out of the plane at O. 
- This happens when current in the wire flows in the negative x-direction. 

Thus, the required current in the wire is 2 A, flowing in the negative x-direction.