Q.A short bar magnet placed with its axis at $30^\circ$ with a uniform external magnetic field of $0.25\ \text{T}$ experiences a torque of magnitude equal to $4.5 \times 10^{-2}\ \text{J}$. What is the magnitude of magnetic moment of the magnet?

Concept & Intuition

The magnetic moment of a planar current loop is defined as $\vec{\mu} = i \, \vec{A}$, where $i$ is the current and $\vec{A}$ is the area vector (magnitude = area, direction perpendicular to the loop). When you have a fixed length of wire, bending it into a circle maximizes the enclosed area for a single turn. The problem gives the wire length $L$ and current $i$; we just need to find the radius of the circle that can be formed, compute its area, and multiply by the current.

Step-by-step reasoning

1. Relate wire length to loop circumference The entire wire of length $L$ is used to make one circular turn. Therefore, the circumference of the loop equals $L$.

2. Compute the area of the loop The area of a circle of radius $r$ is $\pi r^2$. Substituting $r$:

3. Magnetic moment For a single-turn loop carrying current $i$, the magnetic moment is

The concept first

Two long parallel wires a distance $d$ apart, carrying $I_1$ and $I_2$, exert on each other a force per unit length

attractive if the currents are in the same direction, repulsive if opposite ("like currents attract" — the opposite of like charges, a classic memory trap).

Because forces are vectors, we must fix a sign convention. Put the wires on the $x$-axis in the order A, B, C from left to right, and call rightwards positive. Define the convenient unit

Given: $I_A=3i$ (up), $I_B=i$ (down), $I_C=2i$ (up); spacings $AB=BC=d$, so $AC=2d$.

Step-by-step

Part 1 — the net force per unit length on C (this defines $\vec F$).

1. Force from A on C. $I_A$ and $I_C$ are both up ⇒ parallel ⇒ attraction. C is pulled to the left (towards A). Separation $2d$:

2. Force from B on C. $I_B$ is down, $I_C$ is up ⇒ antiparallel ⇒ repulsion. C is pushed away from B, i.e. to the right. Separation $d$:

3. Add:

Part 2 — the force per unit length on A once B is removed.

4. Only wire C remains. $I_A$ and $I_C$ are both up ⇒ attraction, so A is pulled towards C, i.e. rightwards. Separation $2d$:

(This is Newton's third law partner of the $3k$ in step 1 — same magnitude, opposite direction.)

5. Express it in terms of $\vec F$. Since $\vec F = -k$,

So the force on A is three times as large as $\vec F$ and points in the opposite direction — option (D). (Option (B), $3\vec F$, is the trap for anyone who gets the magnitude right but forgets that $\vec F$ points left while the force on A points right.)

ANSWER: D

Setup. The wires sit in the order a — b — c. Given $i_b=i_a$, $i_c=4i_a$, spacing a–b $=d_1$ and b–c $=d_2=2d_1$, so the a–c separation is $d_1+d_2=3d_1$. The force per length between two parallel wires carrying $I_1,I_2$ a distance $r$ apart is

attractive for parallel currents, repulsive for anti-parallel.

Force from b (parallel to a, attractive, toward b).

Force from c (anti-parallel to a, repulsive, away from c).

Net force. $F_{ab}$ pulls a toward b while $F_{ac}$ pushes a away from c — opposite directions — so the magnitudes subtract:

The key idea is to recall a familiar fact: a typical compass needle aligns with the Earth’s field, and that field is weak but measurable — about half a gauss. In SI units, 1 gauss = 10⁻⁴ T, so the surface field is around 0.5 × 10⁻⁴ T = 5 × 10⁻⁵ T. That’s squarely in the 10⁻⁵ T ballpark.

Let’s walk through the reasoning step by step.

1. Recall the typical strength

   The Earth’s magnetic field at the surface is often given as 0.3 to 0.6 gauss. Since 1 gauss = 10⁻⁴ tesla, this converts to $3 \times 10^{-5}$ T to $6 \times 10^{-5}$ T.

2. Identify the order of magnitude

   The order of magnitude is the power of ten when the number is written in scientific notation. Both $3 \times 10^{-5}$ and $6 \times 10^{-5}$ have the exponent –5. So the order is $10^{-5}$ T.

3. Compare with the options

   • (A) $10^{-5}$ T → matches.
   • (B) $10^{-15}$ T → that’s a trillion times weaker (typical of interstellar magnetic fields).
   • (C) $10^{-10}$ T → still 100,000 times weaker (more like a laboratory shielded field).
   • (D) $10^{-20}$ T → absurdly tiny (comparable to fields in deep space between galaxies).

4. Confirm with a sanity check

   A refrigerator magnet is about $10^{-2}$ T (100 gauss). Earth’s field is about 100 times weaker than that, so $10^{-4}$ T is a rough upper bound; the actual value is a bit less, hence $10^{-5}$ T is the correct order.

The core idea here is the Lorentz force on a current-carrying conductor: $\vec{F} = I (\vec{l} \times \vec{B})$, where $\vec{l}$ is a vector along the wire in the direction of the current. The problem gives you the force and the wire's orientation, so you can work backwards to find the magnetic field.

The wire lies along the X-axis, so $\vec{l} = l \hat{i}$. The force is given as $\vec{F} = I B_0 l (\hat{k} - \hat{j})$. Notice the force has no $\hat{i}$ component — that's a crucial clue about which components of $\vec{B}$ can exist.

Let's solve it step by step.

1. Write the cross product explicitly. Let $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$. Then

Since $\hat{i} \times \hat{i} = 0$, $\hat{i} \times \hat{j} = \hat{k}$, and $\hat{i} \times \hat{k} = -\hat{j}$, we get

2. Multiply by current to get force.

The problem states $\vec{F} = I B_0 l (\hat{k} - \hat{j})$. Comparing the two expressions:

• Coefficient of $\hat{k}$: $I l B_y = I l B_0 \implies B_y = B_0$
• Coefficient of $\hat{j}$: $-I l B_z = -I l B_0 \implies B_z = B_0$

3. What about $B_x$?

   The cross product $\hat{i} \times \hat{i} = 0$, so $B_x$ simply does not appear in the force expression. This means $B_x$ can be any value — it is not determined by the force. But the options all have $B_x = B_0$, so we take that.

4. Assemble the field.

   With $B_x = B_0$, $B_y = B_0$, $B_z = B_0$, we have

The question asks for the force acting on an electron moving with velocity $\vec{V}$ in a magnetic field $\vec{B}$. This is a direct application of the Lorentz force law, which describes how charged particles behave in electromagnetic fields.

The key concept is that a magnetic field exerts a force only on a moving charge, and that force is always perpendicular to both the velocity and the magnetic field. This perpendicular nature is captured by the cross product — not the dot product or any division. The dot product would give a scalar (a number), not a vector force, and division of vectors is not a defined operation in physics.

For a particle with charge $q$, the magnetic force is $\vec{F} = q(\vec{v} \times \vec{B})$. For an electron, the charge is $q = -e$, where $e$ is the elementary charge (a positive constant). So the force on an electron is $\vec{F} = -e(\vec{V} \times \vec{B})$. The magnitude is $e|\vec{V} \times \vec{B}|$, and the direction is opposite to that of $\vec{V} \times \vec{B}$.

Now let's examine each option:

1. Option (A): $e(\vec{V} \times \vec{B})$

   This gives the correct magnitude and the correct cross-product form, but the sign is positive. Since the question asks for "force acting on an electron" and gives $e$ as the charge of the electron (a positive constant), the actual force is $-e(\vec{V} \times \vec{B})$. However, in many exam contexts, the magnitude or the expression for the magnitude is what's intended, and the negative sign is understood to come from the charge. Option (A) is the standard textbook expression for the magnitude of the magnetic force on an electron.

2. Option (B): $e(\vec{V} \cdot \vec{B})$

   This is a scalar (dot product), not a vector. Force is a vector, so this cannot be correct. Even if we considered magnitude, the dot product gives the component of velocity along the field, which does not contribute to magnetic force — a charge moving parallel to $\vec{B}$ feels no magnetic force.

3. Option (C): $\dfrac{e\vec{V}}{\vec{B}}$

   Division by a vector is not a defined operation in standard vector algebra. This option is meaningless in physics.

4. Option (D): $\dfrac{e\vec{B}}{\vec{V}}$

   Same issue as (C) — vector division is undefined.