Question

A wire carries a current of 6.52 A in a direction that makes an angle of 44.8 degrees with the direction of a magnetic field of strength 0.258 T. What is the magnetic force on a 3.61 m length of wire

Answer 1

Answer:

$F=4.27N$

Explanation:

From the question we are told that:

Current  $I = 6.52 A$

Angle $\theta= 44.8 \textdegree$

Magnetic field   $B= 0.258 T.$

Length $l=3.61m$

Generally the equation for  Magnetic force F is mathematically given by

 $F=BIlsin\theta$

Therefore

 $F=0.258*6.52*3.61*sin 44.8$

 $F=4.27N$