A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity (g = 10 m/s²) through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire when it travels a vertical distance of 200 m is _____

A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity through a uniform horizontal magnetic field

Q. A $20\,\text{m}$ long uniform copper wire held horizontally is allowed to fall under the gravity $(g = 10\,\text{m/s}^2)$ through a uniform horizontal magnetic field of $0.5$ Gauss perpendicular to the length of the wire. The induced EMF across the wire when it travels a vertical distance of $200\,\text{m}$ is ______ mV.

A. $200\sqrt{10}$

B. $0.2\sqrt{10}$

C. $20\sqrt{10}$

D. $2\sqrt{10}$

Correct Answer: $20\sqrt{10}$

Explanation

When a straight conductor of length $l$ moves with velocity $v$ perpendicular to a magnetic field $B$, the induced EMF is given by

$$ \varepsilon = Blv $$

The wire is released from rest and falls under gravity through a vertical distance of $200\,\text{m}$.

Using the equation of motion,

$$ v^2 = u^2 + 2gh $$

Here,

$$ u = 0,\quad g = 10\,\text{m/s}^2,\quad h = 200\,\text{m} $$

So,

$$ v^2 = 2 \times 10 \times 200 = 4000 $$

$$ v = \sqrt{4000} = 20\sqrt{10}\,\text{m/s} $$

Now convert magnetic field into SI units.

$$ 1\,\text{Gauss} = 10^{-4}\,\text{Tesla} $$

$$ B = 0.5 \times 10^{-4} = 5 \times 10^{-5}\,\text{T} $$

Length of the wire,

$$ l = 20\,\text{m} $$

Substitute into EMF formula,

$$ \varepsilon = (5 \times 10^{-5}) \times 20 \times (20\sqrt{10}) $$

$$ \varepsilon = 2 \times 10^{-2}\sqrt{10}\,\text{V} $$

Converting into millivolts,

$$ \varepsilon = 20\sqrt{10}\,\text{mV} $$

Hence, the induced EMF is

$$ \boxed{20\sqrt{10}\,\text{mV}} $$

Explanation

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