Question

Find the length of the conductor which is moving 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 the conductor.

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Answer 1

Answer:answer is 1 m

Induced EMF =Blvd sin theta

Explanation:

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Answer 2

$\huge{ \underline{ \underline \mathfrak \purple{Required \: Answer}}}$

• The length of the conductor is 1m

Given:-

• Velocity of the conductor, v = 10m/s
• Magnetic field of induction, B = 0.8T
• Electro-motive force (e.m.f) between the ends of the conductor, ε = 8v

To Find:-

• Length of the conductor, I = ?

Solution:-

By using the formula for e.m.f and substitution the value in it, we get :

$\implies \: ε = Blv$

$→8 = 0.8 \times 1 \times 10$

$→ \frac{8}{10} \times 1 \times 10$

$→8 = 8 \times 1$

$→ \frac{8}{8} = 1$

$→1 = I$

Therefore,

• Length of the conductor, I = 1m