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

The angle is
`\theta = 30^o`

Step-by-step 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) [ ( F_b )/(I * l * B ) ]`

substituting values

`\theta =sin^(-1) [ ( 0.18 )/( 2.0 * 0.6 * 0.3 ) ]`

`\theta = 30^o`