Question

11. A wire attached to a 5.8 V battery is in a circuit with a resistance of 18 ohm. A 14 cm length of the wire is in a magnetic field of 0.85 T, and the force on the wire is 22 mN. What is the angle of the wire in the field given that the formula for angled wires in fields is F=ILB(sin θ)?

Answer 1

Answer:

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Answer 2

Answer:

Angle of the wire in the field is $35.45^{o}$

Explanation:

Given emf of a battery, $V= 5.8 V$

       resistance of wire, $R=18\Omega$

       length of the wire, $L=14 cm=0.14m$

            magnetic field, $B= 0.85 T$

       force on the wire, $F= 22 mN=22\times 10^{-3}$

Current through the wire, using Ohm's law,

                         $V=IR\implies I=\frac{V}{R}$

                                                 $=\frac{5.8}{18}=0.32A$

Magnetic force on the wire is given by

                               $F=ILB(sin \theta)$

           $\implies 22\times 10^{-3}=0.32 \times 0.14 \times 0.85 (sin \theta)$

           $\implies 22\times 10^{-3}=0.038 (sin \theta)$

                      $\implies 22=38.08 (sin \theta)$

                   $\implies sin \theta =\frac{22}{38.08} =0.58$

Angle of the wire, $\theta =sin^{-1} 0.58=35.45^{o}$

Angle of the wire in the field is $35.45^{o}$.