Question

A wire carries a current of 0.76 A. This wire makes an angle of 46° with respect to a magnetic field of magnitude 6.0 × 10-5 T. The wire experiences a magnetic force of magnitude 9.4 × 10-5 N. What is the length of the wire?

Answer 1

Answer:

The length of the wire is 2.86 m

Explanation:

Given:

Current $I = 0.76$ A

Magnetic field $B = 6 \times 10^{-5}$ T

Force $F = 9.4 \times 10^{-5}$ N

Angle between wire and magnetic field $\theta =$ 46°

The magnetic force on the current carrying wire is given by,

  $F = BIl\sin \theta$

Where $l =$ length of wire

   $l = \frac{F}{BI \sin 46}$

   $l = \frac{9.4 \times 10^{-5} }{6 \times 10^{-5}\times 0.76 \times 0.7193}$

   $l = 2.86$ m

Therefore, the length of the wire is 2.86 m