The straight wire AB carries a current \(I\). The ends of the wire subtend angles \(\theta_1\) and \(\theta_2\) at the point \(P\) as shown in the figure. The magnetic field at the point \(P\) is:
The problem involves calculating the magnetic field at a point \( P \) due to a current-carrying straight wire, where the ends of the wire make angles \( \alpha \) and \( \beta \) with respect to the point \( P \). The solution can be derived using the Biot-Savart law, which gives the magnetic field generated by a current element.
Step 1: {Understanding the Biot-Savart Law}
The Biot-Savart law provides the magnetic field \( d\mathbf{B} \) at a point due to a small current element \( I \, d\mathbf{l} \). The law is given by: \[ d\mathbf{B} = \frac{\mu_0 I}{4 \pi} \frac{d\mathbf{l} \times \hat{r}}{r^2} \] where: - \( \mu_0 \) is the permeability of free space, - \( I \) is the current, - \( d\mathbf{l} \) is the infinitesimal length of the wire element, - \( \hat{r} \) is the unit vector from the wire element to the point where the magnetic field is being calculated, - \( r \) is the distance from the wire element to the point.
Step 2: {Applying the Biot-Savart Law to a Straight Wire}
For a straight current-carrying wire, the magnetic field at a point \( P \) can be found by integrating the contributions from all infinitesimal elements of the wire. The result for the magnetic field due to a finite straight wire at a point \( P \) is given by: \[ B = \frac{\mu_0 I}{4 \pi d} (\sin \theta_1 - \sin \theta_2) \] where: - \( d \) is the perpendicular distance from the wire to the point \( P \), - \( \theta_1 \) and \( \theta_2 \) are the angles between the line connecting the point \( P \) and the ends of the wire, and the wire itself. The expression is derived by integrating the Biot-Savart law along the length of the wire. The terms \( \sin \theta_1 \) and \( \sin \theta_2 \) come from the geometry of the setup, which involves the angles at which the current elements contribute to the magnetic field.
Step 3: {Conclusion}
Thus, the magnetic field at point \( P \) due to a straight current-carrying wire, where the ends of the wire make angles \( \alpha \) and \( \beta \) with the point, is: \[ B = \frac{\mu_0 I}{4 \pi d} (\sin \theta_1 - \sin \theta_2) \] Therefore, the correct option is (A).
In the given circuit, \(E_1 = E_2 = E_3 = 2V\) and \(R_1 = R_2 = 4\Omega\), then the current flowing through the branch AB is:
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A decimolar solution potassium ferrocyanide is 50% dissociated at 300 K. The osmotic pressure of solution is (R = 8.314 J K$^{-1}$ mol$^{-1}$):