Question

Current through ABC and A′B′C′ is I as shown in the given figure. If PB=PB′=r and C′B′PBC are collinear, the magnetic field at P is:

• A. \(\frac{2I}{4 \pi r} \\[10pt]\)
• B. \(\frac{2 \mu_0 I}{4 \pi r} \\[10pt]\)
• C. \(\frac{\mu_0}{4 \pi r} \\[10pt]\)
• D. Zero

Answer 1

The Correct Option is D

To find the magnetic field at point P, we analyze the contributions from all current-carrying segments:

1. Segment AB: The current flows from A to B. Using the Biot-Savart law, the magnetic field at P (distance r away) would be: \[ B_{AB} = \frac{\mu_0 I}{4\pi r} \] (direction into the page)

2. Segment BC: Since P lies along the collinear extension of BC (C'B'PBC are collinear), the current element and displacement vector are parallel. From Biot-Savart law: \[ dB = \frac{\mu_0 I}{4\pi} \frac{d\vec{l} \times \hat{r}}{r^2} = 0 \] because θ = 0° ⇒ sinθ = 0. Thus, \( B_{BC} = 0 \).

3. Segment APSC: The current flows in the opposite direction (S to A). The magnetic field contribution at P would be: \[ B_{APSC} = -\frac{\mu_0 I}{4\pi r} \] (direction out of the page, opposite to B_AB)

The total magnetic field at P is the sum of these contributions: \[ B_{total} = B_{AB} + B_{BC} + B_{APSC} = \frac{\mu_0 I}{4\pi r} + 0 - \frac{\mu_0 I}{4\pi r} = 0 \]

Final answer: The magnetic field at P is \(\boxed{\text{Zero}}\) (Option 4).

Answer 2

The magnetic field due to a straight current-carrying wire is given by the formula:
\(B = \frac{\mu_0 I}{2 \pi r}\)
where r is the distance from the wire to the point where the magnetic field is being calcu lated. In this case, since the current through ABC and A′B′C′ is equal but flows in opposite directions, the magnetic fields produced at point P by each wire will cancel each other out
Thus, the net magnetic field at point P is zero.