CBSE 2025, Set 3

Question

A metal rod of length $50	ext{ cm}$ is held vertically and moved with a velocity of $10	ext{ m/s}$ towards east. The horizontal component of the Earth's magnetic field at the place is $0.4	ext{ G}$. The emf induced across the ends of the rod is: A). $0.1	ext{ mV}$ B). $0.2	ext{ mV}$ C). $0.8	ext{ mV}$ D). $1.6	ext{ mV}$

Edvaya · Accepted Answer

The correct answer is $0.2	ext{ mV}$. The induced motional emf $\varepsilon$ in a straight conductor moving perpendicular to a magnetic field is given by:

$$\varepsilon = B l v$$

Here, the rod is vertical and moves East. It cuts the horizontal component of the Earth's magnetic field ($B_H$), which points North. Thus, $\vec{v}$, $\vec{B}$, and $\vec{l}$ are mutually perpendicular.

Given values:
* $l = 50	ext{ cm} = 0.5	ext{ m}$
* $v = 10	ext{ m/s}$
* $B = 0.4	ext{ G} = 0.4 	imes 10^{-4}	ext{ T}$

Substitute the values:
$$\varepsilon = (0.4 	imes 10^{-4}) 	imes (0.5) 	imes (10)$$
$$\varepsilon = 2 	imes 10^{-4}	ext{ V}$$
$$\varepsilon = 0.2 	imes 10^{-3}	ext{ V} = \boxed{0.2	ext{ mV}}$$

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Physics > Magnetism Induction And Alternating Current > Electromagnetic Induction

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