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A straight conductor $0.1 \mathrm{m}$ long moves in uniform magnetic field $0.1 \mathrm{T}$. The velocity the conductor is $15 \mathrm{m} / \mathrm{s}$ and is directer perpendicular to the field. The emf inducod between the two ends of the conductor is
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The correct answer is:
$0.15 \mathrm{V}$
Given, length of conductor $l=0.1
\mathrm{m}$
Mangetic field. $B=0.1 \mathrm{T}$
Velocity of conductor, $v=15 \mathrm{m} / \mathrm{s}$
The angle between $\mathbf{v}$ and $\mathbf{B}$ is $90^{\circ}$
When $\mathbf{v}$ and $\mathbf{B}$ are mutually perpendicular, then emf (induced) is given by
$$
\varepsilon=v B l=15 \times 0.1 \times 0.1=\frac{15}{100}=0.15 \mathrm{V}
$$
\mathrm{m}$
Mangetic field. $B=0.1 \mathrm{T}$
Velocity of conductor, $v=15 \mathrm{m} / \mathrm{s}$
The angle between $\mathbf{v}$ and $\mathbf{B}$ is $90^{\circ}$
When $\mathbf{v}$ and $\mathbf{B}$ are mutually perpendicular, then emf (induced) is given by
$$
\varepsilon=v B l=15 \times 0.1 \times 0.1=\frac{15}{100}=0.15 \mathrm{V}
$$
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