Q.A piece of straight wire has mass 20 g and length 1 m. It is to be levitated using a current of 1 A flowing through it and a perpendicular magnetic field B in a horizontal direction. What must be the magnitude of B?

The magnetic force on a current-carrying wire depends on sin(theta) between the current direction and B; when the wire is parallel to B, that force is zero.

The force on a straight current-carrying conductor of length l in a uniform magnetic field B is given by

F = BIl*sin(theta)

where theta is the angle between the direction of current flow and the direction of B. The force is maximum (F = BIl) when the conductor is PERPENDICULAR to B (theta = 90 degrees), and it is ZERO when the conductor is PARALLEL to B (theta = 0 degrees), because sin(0) = 0. Physically, a charge moving exactly along the field line experiences no magnetic deflecting force (F = qv x B is zero when v is parallel to B).

The magnetic force on a current-carrying wire is $F = BIL\sin\theta$, which is maximum when $\sin\theta = 1$, i.e. when the wire is perpendicular to $B$.

The force on a straight wire of length $L$ carrying current $I$ in a uniform field $B$ is

$F = BIL\sin\theta$

where $\theta$ is the angle between the current direction and $B$.

• If the wire is parallel to $B$ ($\theta = 0^\circ$), $\sin\theta = 0$, so $F = 0$.

• If the wire is perpendicular to $B$ ($\theta = 90^\circ$), $\sin\theta = 1$, so $F = BIL$, the maximum possible value.

B has units tesla = Wb/m² = N/(A·m); newton/ampere² is NOT a unit of B.

Magnetic field B can be expressed as:

• tesla (T),
• weber/metre² (Wb/m²), since 1 T = 1 Wb/m²,
• newton/(ampere·metre), from F = BIL ⇒ B = F/(IL) = N/(A·m).

Newton/ampere² does not correspond to the dimensions of magnetic field, so it is the odd one out.

The force on a current-carrying wire in a magnetic field depends on $\sin\theta$, which is maximum (=1) at θ = 90°.

A straight current-carrying conductor of length $L$ carrying current $I$, placed at angle $\theta$ to a uniform magnetic field $B$, experiences a force

$F = BIL\sin\theta$

This is maximum when $\sin\theta$ is maximum, i.e. $\sin\theta = 1$, which occurs at $\theta = 90°$ (wire perpendicular to the field). At $\theta = 0°$ (wire parallel to the field) the force is zero.

Summing the Lorentz force qv x B over all the moving charges in a straight conductor of length l carrying current I gives F = I l x B.

Each charge carrier feels $\vec f=q\vec v\times\vec B$. Summing over all carriers in a segment of length $\vec l$ carrying current $I$ gives $\vec F=I(\vec l \times \vec B)$, with direction given by the right-hand rule.

The net force on any closed current loop in a uniform field is always zero.

For a closed circuit of current $I$ in a uniform magnetic field $\vec{B}$, the total force is

$\vec{F} = I \oint d\vec{l} \times \vec{B} = I\left(\oint d\vec{l}\right)\times \vec{B}$

Since the loop is closed, $\oint d\vec{l} = 0$ (the vector sum of all path elements around a closed loop is zero), so $\vec{F} = 0$. (Note: the net torque need not be zero — that is what causes the loop to rotate.)

Force on a current-carrying wire in a field is $F=Bil\sin\theta$; at $\theta=30^\circ$, $F=iBl/2$.

The magnetic force on a straight wire of length $l$ carrying current $i$ at angle $\theta$ to field $B$ is

$F=Bil\sin\theta.$

With $\theta=30^\circ$, $\sin30^\circ=\tfrac12$, so $F=Bil\times\tfrac12=\dfrac{iBl}{2}.$

1 tesla = 1 weber per square metre, so 5 T = 5 Wb/m².

Magnetic flux density (magnetic field) B has the SI unit tesla. The tesla is defined from magnetic flux Φ = B·A, so B = Φ/A, giving:

$1\ \text{T} = 1\ \frac{\text{Wb}}{\text{m}^2}$

Therefore a field of 5 tesla equals 5 weber per square metre. The powers of ten in the other options are incorrect (there is no factor of 10⁵ or 10²).

Magnetic flux density B measures the strength of a magnetic field, defined either as flux per unit area or as force per unit (current × length).

Magnetic flux density (also simply called the magnetic field, B) at a point can be defined in two equivalent ways:

(1) As flux per unit area: B = dΦ/dA, the magnetic flux passing normally through a unit area held perpendicular to the field at that point.

(2) As force per unit current element: from F = IL×B for a straight current-carrying conductor, B is numerically equal to the force experienced per unit length of a conductor carrying unit current, placed perpendicular to the field.

Its SI unit is the tesla (T), equal to Wb/m² or kg·s⁻²·A⁻¹.