Q.A $3.0\ \text{cm}$ wire carrying a current of $10\ \text{A}$ is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be $0.27\ \text{T}$. What is the magnetic force on the wire?

Concept & Intuition

A long straight wire carrying a steady current produces a magnetic field that circles around it. The strength of this field depends only on the current and the perpendicular distance from the wire — not on the wire’s length, as long as the wire is very long compared to the distance. This is a classic application of Ampère’s law, and the result is beautifully simple: the field falls off as $1/r$, not $1/r^2$ like the electric field from a point charge. That’s because the source (the current) is spread out along a line, not concentrated at a point.

Step-by-step solution

1. Recall the formula for the magnetic field at a perpendicular distance $r$ from an infinitely long straight wire carrying current $I$:

where $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$ is the permeability of free space.

2. Identify the given values:

   • Current, $I = 18\ \text{A}$
   • Distance, $r = 12\ \text{cm} = 0.12\ \text{m}$ (always convert to meters)

3. Substitute into the formula:

4. Simplify step by step:
   • Cancel $\pi$: $\frac{4\pi}{2\pi} = 2$ …

The key concept here is Ampère’s law, which relates the line integral of the magnetic field around a closed loop to the current passing through that loop. For a long, straight, thin-walled pipe, the current flows only on the surface (the wall). Inside the pipe, there is no current. By choosing a circular Amperian loop inside the pipe, coaxial with the pipe, the enclosed current is zero. Ampère’s law then forces the magnetic field to be zero at every point inside.

Let’s work through it step by step.

1. Understand the geometry and current distribution

   The pipe is infinitely long, straight, and thin-walled. This means the current $i$ flows uniformly along the surface of the pipe (like a cylindrical sheet of current). There is no current inside the pipe’s hollow interior.

2. Choose an Amperian loop inside the pipe

   Because of the cylindrical symmetry, the magnetic field inside must be azimuthal (circling around the axis) and depend only on the radial distance $r$ from the axis. We pick a circular loop of radius $r$ (where $r$ is less than the pipe’s inner radius), centered on the axis and lying in a plane perpendicular to the pipe.

3. Apply Ampère’s law

   Ampère’s law states:

For our circular loop, the magnetic field $\vec{B}$ is tangential and constant in magnitude along the loop, so the left side becomes:

The right side, $I_{\text{enc}}$, is the total current passing through the area bounded by the loop. Since the loop lies entirely inside the hollow pipe, no current passes through it — all the current is on the surface, outside the loop. Hence $I_{\text{enc}} = 0$.

4. Conclude the magnetic field inside From Ampère’s law:

This holds for any radius $r$ inside the pipe (including on the axis, where $r=0$). So the magnetic field is zero at every point inside the pipe. …

The concept first

When a conductor moves through a magnetic field, every free charge inside it is carried along at velocity $\vec v$. A moving charge feels the magnetic (Lorentz) force

That force sweeps positive carriers to one end of the rod and leaves the other end negative — a separation of charge, which is exactly what an emf is. Inside the source (the rod), conventional current flows from the low-potential end to the high-potential end, i.e. in the direction of $\vec v\times\vec B$. So the rule to remember is simply:

Also remember the geometry constraint: if $\vec B$ is parallel (or antiparallel) to $\vec v$, the cross product is zero and nothing at all happens. That instantly kills two of the options.

Step-by-step

1. Set up axes. Take $\hat x$ to the right, $\hat y$ up the page, and $\hat z$ out of the page (right-handed set).

2. Write what is given.

3. Impose the physics. We need

4. Eliminate the in-plane options (A) and (B). If $\vec B=\pm B\hat x$, then $\vec v\times\vec B=v\hat x\times(\pm B\hat x)=\vec 0$ — no force on the carriers, no emf, no current. But the figure shows a current, so $\vec B$ cannot lie along the direction of motion. Options (A) and (B) are out.

5. Test "out of the page", $\vec B=B\hat z$: …

Concept — force between two parallel current-carrying conductors.

Step 1 — set up with the given ratio. Let the second conductor carry $I$, so the first carries $2I$. With $L = 50$ m, $d = 0.2$ m, $F = 1$ N:

Step 2 — simplify. …

The key insight here is that the electric field $\vec{E}$ is uniform and directed along the $y$-axis, while the particle's initial velocity $\vec{v}$ is purely along the $x$-axis. This means the force on the particle is $\vec{F} = q\vec{E}$, which is constant in both magnitude and direction (along $y$). There is no force along $x$, so the $x$-component of velocity remains unchanged.

This is exactly analogous to throwing a ball horizontally in a uniform gravitational field — the ball moves with constant horizontal speed while accelerating downward. Here, the electric field plays the role of gravity, and the charge $q$ replaces mass in the force law.

1. Force and acceleration: The electric force is $\vec{F} = qE\,\hat{j}$. By Newton’s second law, the acceleration is $\vec{a} = \frac{qE}{m}\,\hat{j}$, constant and along $y$.

2. Motion along $x$: No force in $x$, so $a_x = 0$. The initial velocity is $v$ along $x$, hence $x(t) = vt$.

3. Motion along $y$: Initial velocity in $y$ is zero. With constant acceleration $a_y = qE/m$, we get $y(t) = \frac{1}{2} \frac{qE}{m} t^2$.

4. Eliminate time: From $x = vt$, we have $t = x/v$. Substituting into $y(t)$:

This is the equation of a parabola: $y \propto x^2$. …

Concept & Intuition

The magnetic field due to a long straight wire circles around it according to the right-hand rule. For a wire carrying current into the page, the field at a point to its right points downward (south), and at a point to its left points upward (north). Here both wires are on the positive x‑axis, and the origin is to their left. So each wire alone produces a field at the origin that points in the same direction (downward, or into the negative y‑direction). Because the fields are vectors in the same direction, the total magnitude is simply the sum of the individual magnitudes. The ratio then becomes a simple comparison of distances.

Step‑by‑step reasoning

1. Magnetic field from a single long wire For an infinitely long straight wire carrying current $i$, the magnitude of the magnetic field at a perpendicular distance $r$ is

The direction is given by the right‑hand rule: if the thumb points in the direction of the current (into the page), the fingers curl in the direction of $\vec{B}$. At a point to the left of the wire, the field points downward (say, the $-\hat{y}$ direction).

2. Field at the origin from the wire at $x = 1$ cm Distance from origin: $r_1 = 1$ cm $= 0.01$ m. Magnitude:

This is exactly $B_0$ (the field if only this wire were present). So

3. Field at the origin from the wire at $x = 2$ cm Distance from origin: $r_2 = 2$ cm $= 0.02$ m. Magnitude: $B_2 = \frac{\mu_0 i}{2\pi (0.02)} = \frac{1}{2} \cdot \frac{\mu_0 i}{2\pi (0.01)} = \frac{1}{2} B_0.$ …

Concept & Intuition

Mutual inductance $M$ between two circuits is defined by $M = \frac{\Phi}{I}$, where $\Phi$ is the magnetic flux through one circuit due to a current $I$ in the other. For a long straight wire, the magnetic field is not uniform — it falls off as $1/r$. So the flux through the rectangular loop must be found by integrating $B$ over the loop’s area. The key geometric detail is that the rectangle’s near side is at distance $a$ from the wire and its far side at distance $2a$ (since the side length along the direction perpendicular to the wire is $a$). This specific ratio gives the clean $\ln 2$ result.

Step-by-step solution

1. Set up coordinates and field expression Place the long straight wire along the $y$-axis. The rectangular frame lies in the same plane (the $xy$-plane). Let the rectangle extend from $x = a$ to $x = 2a$ (so its width perpendicular to the wire is $a$) and have length $b$ parallel to the wire (say from $y=0$ to $y=b$). The current $I$ in the wire produces a magnetic field at a perpendicular distance $x$:

directed into or out of the plane (depending on current direction), but we only care about magnitude for flux.

2. Flux through a thin strip Consider a thin vertical strip of width $dx$ at distance $x$ from the wire, extending the full height $b$ of the rectangle. The area of this strip is $dA = b\,dx$. The magnetic field is constant over this strip (since $x$ is fixed). The flux through the strip is:

3. Integrate over the rectangle’s width …