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

Answer: (D)

Zero

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.