Question

ماهي محصلة كثافة الفيض المغناطيسي عند مركز الحلقة (c)؟

• A. $8 \times 10^{-6} T$
• B. $3.2 \times 10^{-5} T$
• C. $4 \times 10^{-6} T$
• D. $1.6 \times 10^{-5} T$

Answer 1

Answer: (A)

$8 \times 10^{-6} T$

Answer 2

The problem asks for the net magnetic field at the center (c) of a circular metallic loop. The setup involves two sources of magnetic fields:

1. A circular loop with current entering at point x and leaving at point y.
2. A straight wire tangent to the loop, carrying a current of 2A.

1. Magnetic field due to the circular loop:

The current $I$ enters the loop at point x and leaves at point y. This means the current divides into two paths: one through the minor arc (between x and y) and another through the major arc (between x and y).

Since the magnetic fields $B_1$ and $B_2$ have equal magnitudes and opposite directions (one into the page, one out of the page), the net magnetic field due to the circular loop at its center is: $B_{loop} = B_1 - B_2 = 0$.

This is a general result: for a current-carrying loop where current enters and leaves at two points, the net magnetic field at the center is always zero.

2. Magnetic field due to the straight wire:

The straight wire is tangent to the top of the loop and carries a current of $2A$. The average diameter of the loop is 10 cm, so its radius is $r = \frac{10 \text{ cm}}{2} = 5 \text{ cm} = 0.05 \text{ m}$. Since the wire is tangent to the loop, the perpendicular distance from the center 'c' to the straight wire is equal to the radius of the loop, $d = r = 0.05 \text{ m}$.

The magnetic field due to a long straight current-carrying wire is given by: $B_{wire} = \frac{\mu_0 I_{wire}}{2\pi d}$ Given $\mu_0 = 4\pi \times 10^{-7} \text{ Wb/A.m}$ and $I_{wire} = 2 \text{ A}$. $B_{wire} = \frac{(4\pi \times 10^{-7} \text{ Wb/A.m}) \times (2 \text{ A})}{2\pi \times (0.05 \text{ m})} = 8 \times 10^{-6} \text{ T}$.

3. Net magnetic field at center 'c':

The net magnetic field at the center 'c' is the vector sum of the fields due to the loop and the straight wire: $B_{net} = B_{loop} + B_{wire}$ Since $B_{loop} = 0$, $B_{net} = B_{wire} = 8 \times 10^{-6} \text{ T}$.