Q.A current carrying circular loop of radius $R$ is placed in the $x$-$y$ plane with centre at the origin. Half of the loop with $x > 0$ is now bent so that it now lies in the $y$-$z$ plane.

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 $B = 6 \times 10^{-3} \, \mathrm{T}$:

Thus, the speed of the electron is $3 \times 10^6 \, \mathrm{m/s}$.

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.

(A) $B\propto r$ is what holds inside a thick current-carrying conductor, not outside it. (D) a straight line falling to zero at a finite $r$ would mean the field vanishes at a finite distance — false. (C) a hump implies a maximum at some $r>0$ — there is none for an infinitely thin, infinitely long wire.

ANSWER: B

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:

4. Calculate Numerator: $0.250 \times 9.8 = 2.45$ Denominator: $4 \times 2.5 = 10$

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 The wire Q is suspended, meaning the net vertical force on it is zero. The upward magnetic force per unit length equals the weight per unit length:

(Here $m$ is mass per unit length of Q.)

4. Solve for the separation $d$ Rearranging:

5. Match with options The direction is right to left, and the separation is $\frac{25 I \mu_0}{\pi m g}$. This matches option (C).

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).

• Entering: flux into the page through the loop is increasing. The induced current opposes this, so it must produce flux out of the page inside the loop $\Rightarrow$ the induced current flows anticlockwise. With clockwise defined as positive, the emf is negative.
• Leaving: the into-page flux is now decreasing; the induced current tries to maintain it, so it flows clockwise $\Rightarrow$ the emf is positive.

Step 5 — Assemble the graph.

$e = -2\times10^{-4}$ V (rectangular) for $0<t<1$ s; $e=0$ for $1<t<5$ s; $e=+2\times10^{-4}$ V (rectangular) for $5<t<6$ s. That is option (C). (A) and (B) are wrong because they are triangular ramps, and (D) has both pulses with the wrong sign.

ANSWER: C

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