Two identical circular wires P and Q each of radius R and carrying current I are kept in perpendicular planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two coils.
\(\frac{\mu_0 I}{\sqrt{2} R}, \beta = 45^\circ\)
The magnetic field at the centre of a circular current-carrying loop is given by \( B = \frac{\mu_0 I}{2R} \). Since the two coils are perpendicular to each other, their magnetic fields add vectorially.
Using the Pythagorean theorem:
\[ B_{\text{net}} = \sqrt{B_P^2 + B_Q^2} = \sqrt{\left(\frac{\mu_0 I}{2R}\right)^2 + \left(\frac{\mu_0 I}{2R}\right)^2} \]
\[ B_{\text{net}} = \frac{\mu_0 I}{\sqrt{2} R} \]
The angle \( \beta \) of the resultant field with respect to one of the coils is:
\[ \beta = 45^\circ \]
Thus, the correct answer is option 3.
Important Questions from Moving Charge and Magnetism
A square-shaped wire loop of side L is carrying a current I. What is the magnetic field at the point of intersection of diagonals of the square wire loop?
The magnitude of a magnetic force on a current-carrying conductor is given by:
Under the influence of a uniform magnetic field, a charged particle moves with a constant speed v in a circle of radius r. The time period of the revolution of the particle:
A square-shaped wire loop of side L is carrying a current I. What is the magnetic field at the point of intersection of diagonals of the square wire loop?
The magnitude of a magnetic force on a current-carrying conductor is given by:
