Moving Charges and Magnetism — NEET UG practice

1. Since the current is uniformly distributed over the cross-section, we use Ampère's circuital law, $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}$, taking a circular Amperian loop of radius $r$ coaxial with the wire.

2. For a point inside the wire ($r < a$), only the fraction of current enclosed within radius $r$ contributes. Since current density is uniform, the enclosed current is $I \cdot \dfrac{r^2}{a^2}$.

Applying Ampère's law:

$$B(2\pi r) = \mu_0 I \frac{r^2}{a^2}$$

Solving for $B$:

$$B = \frac{\mu_0 I r}{2\pi a^2}$$

This shows that inside the wire, $B$ increases linearly with $r$.

3. For a point outside the wire ($r > a$), the entire current $I$ is enclosed by the Amperian loop.

So,

$$B(2\pi r) = \mu_0 I$$ $$B = \frac{\mu_0 I}{2\pi r}$$

This shows that outside the wire, $B$ decreases as $\dfrac{1}{r}$.

4. Combining both results: $B$ rises linearly from zero at the centre to a maximum at $r = a$, and then falls off as $\dfrac{1}{r}$ for $r > a$.

This behaviour is correctly shown in Option A.

1. Since both arcs $ABC$ and $ADC$ connect the same two points $A$ and $C$, they act like two resistors in parallel across the same potential difference, so the current in each branch is inversely proportional to its resistance.

Given $R_{ABC} = \dfrac{R}{2}$ and $R_{ADC} = R$, applying $I_1 R_{ABC} = I_2 R_{ADC}$:

$$I_1 \cdot \frac{R}{2} = I_2 \cdot R$$

This simplifies to $I_1 = 2I_2$.

2. The two branch currents must add up to the total current supplied to the loop.

So,

$$I_1 + I_2 = I_0$$

Substituting $I_1 = 2I_2$: $2I_2 + I_2 = I_0$, giving $I_2 = \dfrac{I_0}{3}$ and $I_1 = \dfrac{2I_0}{3}$.

3. Each semicircular arc produces a magnetic field at the centre given by $B = \dfrac{\mu_0 I}{4r}$, but the two arcs carry current in opposite senses around the centre, so their fields point in opposite directions.

Hence the net field is the difference of the two individual fields:

$$B_0 = \frac{\mu_0 I_1}{4r} - \frac{\mu_0 I_2}{4r}$$

4. Substituting the values of $I_1$ and $I_2$ found above:

$$B_0 = \frac{\mu_0}{4r}\left(\frac{2I_0}{3} - \frac{I_0}{3}\right) = \frac{\mu_0 I_0}{12r}$$

Hence, the correct answer is Option B.

1. Two parallel current-carrying wires attract each other when their currents are in the same direction. This attractive force is what would try to lift wire $A$ off the floor.

The force per unit length between two long parallel wires separated by distance $h$ is:

$$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi h}$$

2. Here $I_1 = I$ (wire $A$) and $I_2 = 2I$ (wire $B$), so substituting:

$$\frac{F}{l} = \frac{\mu_0 (I)(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}$$

3. For wire $A$ to just stay on the floor (not lift off), this upward magnetic force per unit length must not exceed the weight per unit length of wire $A$, which is $\lambda g$.

So the equilibrium condition is:

$$\frac{\mu_0 I^2}{\pi h} \le \lambda g$$

4. Rearranging for $h$ gives the minimum required height:

$$h \ge \frac{\mu_0 I^2}{\pi \lambda g}$$

Hence, the minimum value of $h$ is $\dfrac{\mu_0 I^2}{\pi \lambda g}$, matching Option C.