Question

An 88.30-cm wire that carries an electric current of 1.35 A perpendicular to the field has experienced a magnetic force of 9.0✕10 -3 N. Solve for the magnitude of the magnetic field that caused this force.

Answer 1

$F=BILsin \theta$

$B= \frac{F}{ILsin \theta}$

$B= \frac{9.0 \times 10^{-3}}{1.35 \times 88.30 \times 10^{-2}sin90}$

$B= \frac{9.0}{1.35 \times 88.30 } \times 10^{-3} \times 10^{2}$

$B= 0.00075 \times 10^{-1}$

$B= 7.5 \times 10^{-6} T$