Question

A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0ms⁻¹, at right angles to the horizontal component of the earth's magnetic field, of 0.3 × 10⁻⁴Wb/m². The value of the induced emf in wire is :

• A. 2.5 × 10⁻³V
• B. 1.1 × 10⁻³
• C. 0.3 × 10⁻³V
• D. 1.5 × 10⁻³V

Answer 1

Answer: (D)

1.5 × 10⁻³V

Answer 2

The induced electromotive force (emf) in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by the formula:

$e = B l v \sin\theta$

where $\theta$ is the angle between the velocity vector $\vec{v}$ and the magnetic field vector $\vec{B}$.

In this problem:

1. Length of the wire, $l = 10 \text{ m}$.
2. Speed of the wire, $v = 5.0 \text{ m/s}$. The wire is falling, so its velocity vector is directed downwards.
3. Magnitude of the horizontal component of the Earth's magnetic field, $B = 0.3 \times 10^{-4} \text{ Wb/m}^2$. This magnetic field vector is in the horizontal plane.

The problem states that the wire is falling "at right angles to the horizontal component of the earth's magnetic field".

Since the velocity vector $\vec{v}$ is vertically downwards and the magnetic field vector $\vec{B}$ (horizontal component) is in the horizontal plane, $\vec{v}$ is always perpendicular to $\vec{B}$. Therefore, the angle $\theta$ between $\vec{v}$ and $\vec{B}$ is $90^\circ$, which means $\sin\theta = \sin 90^\circ = 1$.

Now, substitute the values into the formula:

$e = (0.3 \times 10^{-4} \text{ Wb/m}^2) \times (10 \text{ m}) \times (5.0 \text{ m/s}) \times 1$

$e = 0.3 \times 10^{-4} \times 50 \text{ V}$

$e = 15 \times 10^{-4} \text{ V}$

$e = 1.5 \times 10^{-3} \text{ V}$