A wire of length $50 c m$ moves with a velocity of $300 m / m i n$ , perpendicular to a magnetic field. If the $e . m . f$ . induced in the wire is $2 V$ , the magnitude of the field in tesla is

$2$

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$5$

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$0.8$

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$2.5$

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Solution

The correct option is C
$0.8$

Step 1. Given data.
Length of the wire $l = 50 c m = 0.5 m$
Velocity is $v = 300 m / m i n = \frac{300}{60} = 5 \frac{m}{s}$
Induced $e . m . f$ is $e = 2 V$
Step 2. Calculating magnetic field
We know the magnetic field is given as $B = \frac{e}{l v}$
Where, $B$ is the magnetic field
$e$ is the induced e.m.f.
$l$ is the length of the wire
$v$ is the velocity
Now, by substituting the given values we get
$B = \frac{2}{0.5 \times 5} = \frac{2}{2.5} = 0.8 T$
Hence, the correct option is, (C).

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A wire of length 50cm moves with a velocity of 300m/min, perpendicular to a magnetic field. If thee.m.f. induced in the wire is 2V, the magnitude of the field in tesla is

A wire of length $50 c m$ moves with a velocity of $300 m / m i n$ , perpendicular to a magnetic field. If the $e . m . f$ . induced in the wire is $2 V$ , the magnitude of the field in tesla is