Chapter 24, Problem 11STP

To determine

The angle of the current carrying wire with the magnetic field.

Answer to Problem 11STP

The angle made by the wire with the field is $\theta =35.28°$.

Explanation of Solution

Given:

A wire, 14 cm long, having a resistance of 18 Ω is attached to a 5.8-V battery. It is kept in a uniform magnetic field of strength 0.85 T, so that it experiences a force of 22 mN.

Formula Used:

The force acting on a wire carrying current is dependent on the strength of magnetic field, the current, length of the wire and the angle which the wire makes with the magnetic field. It is mathematically expressed as,

  $F=ILB\left(\sin \theta \right)$

From this, the angle of inclination can be found as,

  $\begin{array}{l}\sin \theta =\frac{F}{ILB}\\ \theta ={\sin }^{-1}\left(\frac{F}{ILB}\right)\end{array}$ …… (1)

Here, when the wire with a particular resistance is connected to a given voltage supply, the current in the wire can be calculated using the Ohm’s law as,

  $\begin{array}{l}V=IR\\ I=\frac{V}{R}\end{array}$

Calculation:

Here, the current in the wire can be calculated using the supply voltage $V=5.8-\text{V}$ and resistance of the wire $R=18\Omega$ as,

  $\begin{array}{c}I=\frac{5.8}{18}\\ =0.32\text{ A}\end{array}$

Now, for the given strength of magnetic field $B=0.\text{85 T}$ , current $I=0.32\text{ A}$ , length $L=14\text{ cm}$ and the force experienced by the wire $F=22\text{ mN}$ , the angle of inclination of the wire with the magnetic field , $\theta$ can be calculated using (1) as,

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{F}{ILB}\right)\\ ={\sin }^{-1}\left(\frac{22\times {10}^{-3}}{0.32\times 14\times {10}^{-2}\times 0.85}\right)\\ ={\sin }^{-1}\left(0.5777\right)\\ =35.28°\end{array}$

Conclusion:

Therefore, angle of inclination of the current carrying wire with the magnetic field is $35.28°$ .