# A wire of length 0.680 m carries a current of 21.0 A. In the presence of a 0.470 T magnetic field, the wire experiences a force of 5.41 N. What is the angle (less than 90°) between the wire and the magnetic field?

Jul 2, 2016

$\alpha = {53.13}^{\circ}$

#### Explanation:

Given

• $L \to \text{Length of the current carrying wire} = 0.680 m$

• $B \to \text{Applied magnetic field} = 0.47 T$

• $I \to \text{Current passing through wire} = 21 A$

• $F \to \text{Force experienced by the wire} = 5.41 N$

• alpha->"Angle between the wire and magnetic field"=?

Working Formula

• $\vec{F} = \vec{I L} \times \vec{B} = I L B \sin \alpha$

Putting the given values

$5.41 = 21 \times 0.680 \times 0.47 \times \sin \alpha$

$\implies \sin \alpha = \frac{5.41}{21 \times 0.680 \times 0.47} \approx 0.8$

$\alpha = {53.13}^{\circ}$