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diamong [38]
3 years ago
6

a straight wire that is 0.60m long and carrying a current of 2.0a is placed at an angle with respect to a magnetic field of stre

nght 0.30t if the wire experiences a force of magnitude 0.18 n what angle does the wire
Physics
1 answer:
Rashid [163]3 years ago
8 0

Answer:

The  angle is  \theta =  30^o

Explanation:

From the question we are told that

    The length of the wire is  l = 0.60 \ m

    The current is  I  =  2.0 \ A

    The magnetic field strength is  B  =  0.30 \ T

     The magnitude of the magnetic force is  F _b  =  0.18  \ N

Generally the magnetic force exerted on the wire is mathematically represented as

      F_b  =  I  *  l  *  B * sin \theta

Making \theta the subject

      \theta =sin^{-1} [ \frac{ F_b }{I  *  l  *  B } ]

substituting values    

      \theta =sin^{-1} [ \frac{ 0.18 }{ 2.0   *   0.6  *   0.3 } ]

     \theta =  30^o

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A wire of resistance R is cut into ten equal parts which are then connected in parallel. The equivalent resistance of the combin
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Answer:

<em>The equivalent resistance of the combination is R/100</em>

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<u>Electric Resistance</u>

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

Parallel connection of resistances: If R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

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If we have 10 wires of resistance R/10 each and connect them in parallel, the equivalent resistance is:

\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}...+\frac{1}{R/10}

This sum is repeated 10 times. Operating each term:

\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}

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