Q.An electric dipole with dipole moment $4 \times 10^{-9}\,\text{C m}$ is aligned at $30^\circ$ with the direction of a uniform electric field of magnitude $5 \times 10^{4}\,\text{N C}^{-1}$. Calculate the magnitude of the torque acting on the dipole.

The electric dipole moment is a vector that measures the strength and orientation of a dipole. Its magnitude is simply $p = qd$, where $q$ is the magnitude of one charge and $d$ is the separation between the charges. The direction is always from the negative charge toward the positive charge — this is the standard convention in physics.

Let’s apply this step by step.

1. Identify the given quantities

   • Charge magnitude: $q = 0.01\ \text{C}$
   • Separation distance: $d = 0.2\ \text{mm} = 0.2 \times 10^{-3}\ \text{m} = 2 \times 10^{-4}\ \text{m}$
   • Charges are opposite in sign, so this is a pure dipole.

2. Compute the magnitude of the dipole moment

3. Determine the direction

   By definition, the dipole moment vector points from the negative charge to the positive charge. The problem states the charges are opposite in sign, so the direction is along the line joining them, from negative to positive.

4. Match with the options

   • Option (A): magnitude $2 \times 10^{-3}$ — too large. …

Step 1 — Fix the geometry from the wording.

The two charges are $2R$ apart. Sphere B has radius $2R$ and passes through the charge $+q$ — a sphere of radius $2R$ whose surface contains $+q$ must be centred on the other charge, $-q$ (the only point at distance $2R$ from $+q$ on the dipole axis). Sphere A, radius $R$, is drawn about that same centre, and indeed a sphere of radius $R$ centred at $-q$ has its surface passing through the mid-point of the dipole, exactly as stated.

Step 2 — Find the charge enclosed by A.

Sphere A is centred on $-q$ and has radius $R$, while $+q$ lies at a distance $2R > R$ from that centre. Hence:

$q_{\text{enclosed}} = -q \qquad (+q \text{ is outside A})$ …

The key to this question is recognizing that joule per tesla is not a common way to write a unit, but it can be simplified by recalling the relationship between energy, magnetic field, and magnetic moment. The magnetic dipole moment $\vec{\mu}$ of a current loop is defined as $\vec{\mu} = I \cdot \vec{A}$, where $I$ is current (amperes) and $A$ is area (m²). Its SI unit is A·m². Meanwhile, the energy of a magnetic dipole in a uniform magnetic field $B$ is $U = -\vec{\mu} \cdot \vec{B}$, so energy (joules) = magnetic moment (A·m²) × magnetic field (tesla). Rearranging, magnetic moment = energy / magnetic field = J/T. So J/T is exactly the unit of magnetic dipole moment.

Let’s verify each option:

1. Magnetic permeability ($\mu$) has units of henry per meter (H/m) or, equivalently, N/A². Since 1 H = 1 J/A², we get J/(A²·m), not J/T. So (A) is wrong.

2. Magnetic induction (magnetic flux density $B$) is measured in tesla (T) itself. 1 T = 1 N/(A·m) = 1 J/(A·m²). That is not J/T (which would be J per tesla, not tesla itself). So (B) is wrong.

3. Magnetic susceptibility ($\chi$) is a dimensionless quantity (ratio of magnetization to applied field). It has no units at all, so certainly not J/T. So (C) is wrong. …

The core idea here is that torque on a magnetic dipole depends only on the dipole moment $M$, the external field $B$, and the angle between them. Since the bar magnet and the solenoid are said to have the same magnetic moment, we can first find $M$ from the torque data, then use the formula for a solenoid’s magnetic moment ($M = NIA$) to solve for the current $I$.

1. Recall the torque on a magnetic dipole

A magnetic dipole of moment $M$ placed in a uniform external field $B$ experiences a torque

where $\theta$ is the angle between the dipole axis and the field direction.

Here $\theta = 45^\circ$, $B = 400 \times 10^{-4}\,\text{T}$, and $\tau = 0.024\,\text{Nm}$.

2. Solve for the magnetic moment $M$

Since $\sin 45^\circ = \frac{1}{\sqrt{2}}$,

3. Magnetic moment of a solenoid

A solenoid with $N$ turns, cross-sectional area $A$, and carrying current $I$ has magnetic moment

$M = N I A$ …

The key idea is that the torque on an electric dipole in a uniform electric field is given by $\vec{\tau} = \vec{p} \times \vec{E}$, so its magnitude is $|\vec{\tau}| = pE \sin\theta$. The maximum torque occurs when $\sin\theta = 1$, i.e., when the dipole is perpendicular to the field. Therefore, the maximum torque is simply $pE$ — no angles, no fuss. The problem reduces to comparing two simple products.

1. Identify the given data

   Dipole 1: $p_1 = 3.9 \times 10^{-30}\ \mathrm{C\,m}$, $E_1 = 16 \times 10^{4}\ \mathrm{N/C}$

   Dipole 2: $p_2 = 5.2 \times 10^{-30}\ \mathrm{C\,m}$, $E_2 = 4 \times 10^{4}\ \mathrm{N/C}$

2. Write the expression for maximum torque

   For any dipole, $\tau_{\max} = pE$. So:

3. Compute the ratio

Cancel the common factor $10^{-30} \times 10^{4} = 10^{-26}$:

4. Simplify the numbers First, $16/4 = 4$, so: $= \frac{3.9 \times 4}{5.2} = \frac{15.6}{5.2}$ …

[!TLDR]

Halving the torque means the magnet-field angle changes from $90^\circ$ to $30^\circ$, so the required rotation is $60^\circ$.

Concept

A magnetic dipole of moment $m$ in a uniform field $B$ experiences a couple $\tau = mB\sin\theta$ (CBSE Class 12 magnetism), where $\theta$ is the angle between the dipole axis and the field. The torque is largest when the magnet is perpendicular to the field.

Solution

Initially the magnet is perpendicular to the field, so $\theta = 90^\circ$ and

We need the couple halved:

With $\tau_2 = mB\sin\theta_2$:

$mB\sin\theta_2 = \frac{mB}{2} \;\Rightarrow\; \sin\theta_2 = \frac12 \;\Rightarrow\; \theta_2 = 30^\circ.$ …

1. The concept — resolving Earth's field.

Earth's magnetic field $\vec{B}$ at a place points along the direction of a freely suspended magnetic needle. The angle of dip $\delta$ is the angle this resultant field makes with the horizontal. Resolving $\vec{B}$ into components:

2. Why this formula applies.

We are told the horizontal component and the dip, and asked for the resultant. That is exactly the pair connected by $B_H = B\cos\delta$ — no other data is needed.

3. Substitute.

$B = 3\sqrt{2}\times 10^{-5}\,\text{T}$ …

1. The concept — polar vs non-polar molecules.

• In a non-polar molecule (e.g. $\mathrm{O_2}$, $\mathrm{CH_4}$) the centres of positive and negative charge coincide in the absence of an external field. The dipole moment is zero; a field can only induce a temporary dipole.
• In a polar molecule (e.g. HCl, $\mathrm{H_2O}$) the two centres are permanently separated, because the bonded atoms have different electronegativities. The molecule carries a permanent dipole moment $\vec{p} = q\vec{d}$ even with no field applied.

2. Test each statement.

• (A) "Centres of positive and negative charges are separated in the absence of an external field." — TRUE. This is precisely what makes a molecule polar.

• (B) "Centres … are separated in the presence of an external field." — TRUE. They were already separated; an external field additionally aligns the dipole (and slightly increases the separation). The statement is not false. …

Step 1 — The concept: line density measures field strength. In a field-line diagram, the magnitude of $\vec{E}$ is proportional to the number of lines crossing unit area, i.e. inversely proportional to the separation between adjacent lines:

Where lines crowd together the field is strong; where they spread out the field is weak.

Step 2 — Compare A and B. From the figure, the separation on the left (at B) is twice that on the right (at A):

With $E_A = 40\ \text{V m}^{-1}$,

Step 3 — Force on the charge at B. The force on a charge $q$ in a field $E$ is

Convert the charge: $q = 20\ \mu\text{C} = 20 \times 10^{-6}\ \text{C}$. Hence …

For an infinitely long straight wire with linear charge density $\lambda$, the field at distance $r$ is

$E = \dfrac{\lambda}{2\pi\varepsilon_0\,r} = \dfrac{2k\lambda}{r},\qquad k = 9\times 10^{9}\ \text{N m}^2\text{C}^{-2}.$ …

Step 1 — The concept.

Faraday's law of electromagnetic induction states that the emf induced in a coil equals the negative time-rate of change of the magnetic flux linked with it:

The minus sign is Lenz's law (it fixes the direction); the question asks only for the magnitude, so we take $|d\phi/dt|$. Notice that the constant term $9$ in $\phi$ contributes nothing — only changing flux induces an emf.

Step 2 — Differentiate.

$\phi = 3t^2 + 4t + 9$ …