what is the length of conductor which moves with a speed of 10 m/s in the direction perpendicular to the direction of magnetic field of induction 0.8T if it induces an emf of 8v between the ends of conductor
Answers
Answer:
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Explanation:
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Explanation:
Explanation:
Given :
A speed of 10 m/s in the direction perpendicular .
the direction of magnetic field of induction 0.8T
if it induces an emf of 8v between the ends of conductor
To Find :
what is the length of conductor which moves
Solution :
Concept :
The formula of effective EMF and effective internal
So, V = ε – Ir,
where V is the potential difference across the circuit, ε is the emf, I is the current flowing through the circuit, r is internal resistance.
______________________
\begin{gathered}: \implies \sf \: \: \: \: \: \: \: \: \xi = blv \\ \\ \\ : \implies \sf \: \: \: \: \: \: \: \:l = \frac{ \xi}{bv} \\ \\ \\\end{gathered}
:⟹ξ=blv
:⟹l=
bv
ξ
Substitute all values :
\begin{gathered}: \implies \sf \: \: \: \: \: \: \: \: l = \frac{8}{0.8 \times 10} \\ \\ \\ : \implies \sf \: \: \: \: \: \: \: \: l = \cancel{\frac{8}{8}} \\ \\ \\ : \implies \sf \: \: \: \: \: \: \: \: l =1 m\end{gathered}
:⟹l=
0.8×10
8
:⟹l=
8
8
:⟹l=1m
Hence the value of length is 1 m