Question

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 strenght 0.30t if the wire experiences a force of magnitude 0.18 n what angle does the wire

Answer 1

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    

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

     $\theta = 30^o$