Q.A conducting thick copper rod of length 1 m carries a current of 15 A and is located on the Earth's equator, where the magnetic flux lines of the Earth's magnetic field are horizontal, directed from south to north, with a field of $1.3\times10^{-4}$ T. What are the magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east? (A) $14\times10^{-4}$ N, downward (B) $20\times10^{-4}$ N, downward (C) $14\times10^{-4}$ N, upward (D) $20\times10^{-4}$ N, upward

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)

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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 $F = 0$. …

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). …

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

$F = BIL\sin\theta$

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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.

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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

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

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

$F=Bil\sin\theta.$ …

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:

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

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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.

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