Electromagnetic Induction and Alternating Current 7 Question 1

1. A solid metal cube of edge length $2 cm$ is moving in a positive $Y$-direction at a constant speed of $6 m / s$. There is a uniform magnetic field of $0.1 T$ in the positive $Z$-direction. The potential difference between the two faces of the cube perpendicular to the $X$-axis is

(Main 2019, 10 Jan I)

(a) $2 mV$

(b) $12 mV$

(c) $6 mV$

(d) $1 mV$

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Answer:

Correct Answer: 1. (b)

Solution:

  1. Potential difference between opposite faces of cube is

$$ V=\text { induced emf }=B l v $$

where,

$B=$ magnetic field $=0.1 T$,

$l=$ distance between opposite faces of cube

$=2 cm=2 \times 10^{-2} m$ and $v=$ speed of cube $=6 ms^{-1}$.

Hence, $\quad V=0.1 \times 2 \times 10^{-2} \times 6=12 mV$



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