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

Application of Biot-Savart and Ampere's Laws
- The magnetic field at the center \( O \) of the circular loop is calculated as: \[ B_{\text{loop}} = \frac{\mu_0 I_{\text{loop}}}{2R} \]. Given \( I_{\text{loop}} = 1.0 A \) and \( R = 10 \) cm (0.1 m), the calculation is: \[ B_{\text{loop}} = \frac{(4\pi \times 10^{-7}) (1)}{2 \times 0.1} \] which results in \[ B_{\text{loop}} = 2 \times 10^{-6} \text{ T} \].
- The magnetic field generated by a long straight wire at a distance \( d = 20 \) cm is given by: \[ B_{\text{wire}} = \frac{\mu_0 I_{\text{wire}}}{2\pi d} \]. Substituting the values: \[ B_{\text{wire}} = \frac{(4\pi \times 10^{-7}) I_{\text{wire}}}{2\pi \times 0.2} \]. This simplifies to: \[ B_{\text{wire}} = \frac{2 \times 10^{-7} I_{\text{wire}}}{0.2} \] and further to \[ B_{\text{wire}} = 10^{-6} I_{\text{wire}} \].
- For the net magnetic field at \( O \) to be zero, the magnitudes of the magnetic fields must be equal: \[ B_{\text{loop}} = B_{\text{wire}} \]. Therefore, \[ 2 \times 10^{-6} = 10^{-6} I_{\text{wire}} \], yielding \( I_{\text{wire}} = 2 A \). Direction of Current in the Wire:
- Applying the Right-Hand Rule, the magnetic field from the circular loop at point \( O \) is directed into the plane.
- To achieve a net magnetic field of zero, the wire must produce a magnetic field directed out of the plane at \( O \).
- This requires the current in the wire to flow in the negative x-direction.

Consequently, the wire requires a current of 2 A, directed along the negative x-axis.