Q.The dimensional formula of the magnetic field intensity is :

The key here is to look past the two "global" effects (net force and torque) — both of which vanish in this configuration — and examine the force on each individual element of the loop.

1. Set up the geometry. The loop lies in the $x$–$y$ plane and the field is $\vec{B} = B\hat{k}$ (along the $z$-axis). The current $I$ flows clockwise as seen from the $+z$ direction, so by the right-hand rule the loop's magnetic moment $\vec{m} = IA\hat{n}$ points along $-\hat{k}$, anti-parallel to $\vec{B}$.

2. No translation. For any closed loop in a uniform field the net force is

because $\oint d\vec{l} = 0$ around a closed path. This immediately rules out options (A) and (B) — the loop cannot move along the $x$- or $y$-axis.

3. No rotation. The torque is $\vec{\tau} = \vec{m}\times\vec{B}$. Here $\vec{m}$ is anti-parallel to $\vec{B}$, so $\vec{\tau} = 0$ — the loop does not turn.

4. Force on each element — the deciding step. Take the element of the loop at the point $(R, 0, 0)$. For a clockwise current (seen from $+z$), the tangent there points along $-\hat{j}$, so $d\vec{l} = -dl\,\hat{j}$ and

which points along $-\hat{i}$ — i.e. radially inward, toward the centre of the loop. By the symmetry of the circle the same is true at every point: each element is pushed straight toward the centre. …

The key idea is that a magnetic field exerts a force on a moving charge, and a current-carrying wire is just a collection of moving charges. The force on a straight wire of length $\vec{L}$ (a vector pointing in the direction of current) in a uniform magnetic field $\vec{B}$ is $\vec{F} = I (\vec{L} \times \vec{B})$. This is a cross product, so only the component of $\vec{B}$ perpendicular to the wire matters.

Let’s work through it step by step.

1. Identify the vector length of the wire. The wire is $10\ \text{cm} = 0.10\ \text{m}$ long, lying along the $y$-axis, with current in the positive $y$-direction. So the length vector is:

2. Write the magnetic field in SI units. The field is given as $\vec{B} = (5\ \text{mT})\hat{j} - (8\ \text{mT})\hat{k}$. Since $1\ \text{mT} = 10^{-3}\ \text{T}$, we have:

3. Apply the force formula. The current is $I = 1.0\ \text{A}$. So:

Compute the cross product term by term. Remember:

• $\hat{j} \times \hat{j} = 0$ (parallel vectors give zero cross product)
• $\hat{j} \times \hat{k} = \hat{i}$ (right-hand rule: $y$ cross $z$ gives $x$)

So:

Therefore:

Multiplying by $I = 1.0\ \text{A}$:

$\vec{F} = -8\times10^{-4}\,\hat{i}\ \text{N}$ …

The key insight here is a beautiful simplification: in a uniform magnetic field, the net magnetic force on any current-carrying wire segment depends only on the vector displacement between its endpoints, not on the path the wire takes between them. This is because the force on a small element $d\vec{l}$ is $I\,d\vec{l} \times \vec{B}$, and when $\vec{B}$ is constant, the integral $\int d\vec{l}$ over the wire is just the straight-line vector from start to end.

Let's apply this to the triangular loop.

1. Set up the geometry. The loop PQR is an isosceles right triangle with the right angle at P. So PR and PQ are the perpendicular legs, and QR is the hypotenuse. The uniform magnetic field $\vec{B}$ points along PR. Let's assign directions: take PR along the $+y$ axis, and PQ along the $+x$ axis. Then the current direction matters — we need to be consistent. The loop is closed, so current flows P → Q → R → P (or the reverse; the magnitude of force is unaffected by sign, but direction matters for comparing forces).

2. Force on arm PQ. Arm PQ is a straight wire of length $L$ (say) along the $x$-axis. The current in PQ flows from P to Q, so $d\vec{l}$ is along $+\hat{x}$. The magnetic field is $\vec{B} = B\,\hat{y}$. The force on a straight wire is $\vec{F}_{PQ} = I\,(\vec{L}_{PQ} \times \vec{B})$, where $\vec{L}_{PQ}$ is the vector from P to Q (length $L$, direction $+\hat{x}$).

   Compute: $\vec{L}_{PQ} \times \vec{B} = (L\hat{x}) \times (B\hat{y}) = LB\,(\hat{x} \times \hat{y}) = LB\,\hat{z}$.

   So $\vec{F}_{PQ} = I L B \,\hat{z}$. The magnitude is $F = I L B$, and it points out of the plane (say upward). The problem states this force is $F$, so $F = I L B$.

3. Force on arm QR. Arm QR is the hypotenuse. Its vector from Q to R: Q is at $(L,0)$, R is at $(0,L)$ (since PR = PQ = L for an isosceles right triangle). So $\vec{L}_{QR} = \vec{R} - \vec{Q} = (0 - L)\hat{x} + (L - 0)\hat{y} = -L\hat{x} + L\hat{y}$.

   The force: $\vec{F}_{QR} = I\,(\vec{L}_{QR} \times \vec{B}) = I\,[(-L\hat{x} + L\hat{y}) \times (B\hat{y})]$.

   Compute the cross product term by term:

   • $(-L\hat{x}) \times (B\hat{y}) = -L B\,(\hat{x} \times \hat{y}) = -L B\,\hat{z}$.
   • $(L\hat{y}) \times (B\hat{y}) = L B\,(\hat{y} \times \hat{y}) = 0$. So $\vec{F}_{QR} = -I L B \,\hat{z} = -F\,\hat{z}$. …