Q.(a) A circular coil of $30$ turns and radius $8.0\ \text{cm}$ carrying a current of $6.0\ \text{A}$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0\ \text{T}$. The field lines make an angle of $60^\circ$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

The key concept here is the velocity selector — a setup where perpendicular electric and magnetic fields are applied so that only particles with a specific speed pass straight through without deflection. For an electron (or any charged particle), the electric force is $F_E = qE$ and the magnetic force is $F_B = qvB$ (when $\vec{v} \perp \vec{B}$). For the path to remain undeviated, these forces must cancel exactly, meaning they are equal in magnitude and opposite in direction. This gives a simple relation that lets us solve for the speed directly.

1. Set up the force balance condition. The electric force on the electron is $F_E = eE$ (where $e$ is the magnitude of the electron’s charge), directed opposite to the electric field because the electron is negatively charged. The magnetic force is $F_B = evB$ (since $\vec{v} \perp \vec{B}$), and its direction is given by the right-hand rule (or left-hand rule for negative charge). For the beam to go straight, we require:

The charge $e$ cancels out, leaving:

2. Solve for the speed $v$. Rearranging:

3. Substitute the given values. $E = 1.8 \times 10^4 \, \mathrm{V/m}$ and $B = 6 \times 10^{-3} \, \mathrm{T}$: $$$$

Step 1 — Apply Ampère's circuital law.

Take a circular Amperian loop of radius $r$ centred on the wire and lying in the plane perpendicular to it. By symmetry $\vec B$ has the same magnitude everywhere on that circle and is tangential to it, so

(The direction is given by the right-hand thumb rule, but only the magnitude matters for the graph.)

Step 2 — Shape of $B$ versus $r$.

• Monotonically decreasing, so any rising graph is out.
• $B\to\infty$ as $r\to 0$ (the curve hugs the $B$-axis) and $B\to 0$ only as $r\to\infty$ — so it approaches the $r$-axis asymptotically and never crosses it.
• This is a rectangular hyperbola ($Br=\text{const}$), concave up.

Step 3 — Eliminate the distractors. …

Concept & Intuition

The wire is “suspended in mid air” — that means it’s not falling, so the net vertical force is zero. The only two vertical forces are its weight (downward) and the magnetic force (upward). For a current-carrying wire in a uniform magnetic field, the magnetic force is given by $F = BIL\sin\theta$. Here the field is horizontal and the wire is straight; to get an upward force, the current direction must be perpendicular to the field. The problem states the field is “uniform horizontal” — so if the wire is horizontal and the current flows perpendicular to the field, the force is vertical. We set that equal to the weight.

Step-by-step

1. Identify the forces

   Weight: $W = mg$, where $m = 250\ \text{g} = 0.250\ \text{kg}$, $g = 9.8\ \text{m/s}^2$.

   Magnetic force: $F_m = B I L \sin\theta$. For maximum upward force, we take $\theta = 90^\circ$ so $\sin\theta = 1$. Thus $F_m = B I L$.

2. Set up the equilibrium condition

   Suspended means $F_m = W$:

3. Solve for $B$

Substitute values: …

Concept & Intuition

Two parallel current-carrying wires exert a magnetic force on each other: currents in the same direction attract, opposite directions repel. Here wire Q is above P and must be held up against gravity. The only upward force available is the magnetic force from P. Since gravity pulls Q downward, the magnetic force must be upward — that means the wires must attract each other. Attraction occurs when the currents are in the same direction. But careful: P’s current is left to right; if Q’s current is also left to right, they attract, pulling Q downward (toward P). That would add to gravity, not oppose it. So we need the opposite: Q’s current must be right to left, making the currents antiparallel, which gives a repulsive force — pushing Q upward. That repulsion can balance weight.

Now, the magnitude of the magnetic force per unit length between two long parallel wires is

where $d$ is the separation. For a wire of mass $m$ (presumably per unit length, or total mass — the problem’s phrasing implies $m$ is the mass per unit length, as the force per unit length is what matters for suspension), the weight per unit length is $mg$. Setting repulsive force upward equal to weight downward gives the equilibrium separation.

Step-by-step solution

1. Identify the force direction

   Wire P carries current $50\,\text{A}$ left to right. For wire Q to be repelled upward, its current must be opposite to P’s — i.e., right to left. This eliminates option (B) (which says left to right).

2. Write the magnetic force per unit length

   For two long parallel wires with currents $I_1$ and $I_2$ separated by distance $d$, the force per unit length is

Here $I_1 = 50\,\text{A}$, $I_2 = I$ (the current in Q). So

3. Set up equilibrium …

Step 1 — Timeline of the motion.

Speed $v = 2$ cm s$^{-1}$, loop side $l = 2$ cm, field region width $= 10$ cm.

• Entry: the front edge is in the field but the rear edge is not, for the time the loop needs to advance its own length: $t = 2/2 = 1$ s. So $0 \le t \le 1$ s.
• Fully inside: front edge reaches the far boundary when it has travelled 10 cm, i.e. at $t = 10/2 = 5$ s. So from $t=1$ s to $t=5$ s the loop is entirely inside.
• Exit: takes another $1$ s, from $t=5$ s to $t=6$ s.

Step 2 — Magnitude of the emf.

While entering or leaving, only one side of length $l$ cuts field lines (the other is outside the field), so

Because $B$, $l$ and $v$ are all constant, this emf is constant throughout each 1 s interval — the pulses are rectangular, not triangular. (Equivalently, $\Phi = B\,l\,(vt)$ increases linearly, so $e=-d\Phi/dt$ is constant.)

Step 3 — When the loop is fully inside (1 s $\to$ 5 s).

The flux $\Phi = B\,l^2$ is now constant (both the entering and leaving edges are inside the same uniform field), so $d\Phi/dt = 0$ and $e = 0$.

Step 4 — Sign, by Lenz's law (clockwise taken positive). …

In crossed electric and magnetic fields, the beam passes straight through when the electric and magnetic forces cancel: …