Chapter 26, Problem 23P

To determine

The magnetic field.

Answer to Problem 23P

  $\left(10\widehat{i}+10\widehat{j}-15\widehat{k}\right)\text{T}$

Explanation of Solution

Given:

Equilibrium angular displacement of the wire from vertical $=\theta$

Magnitude of current in the wire $=I=2\text{ A}$

Length of the wire before rotating $=\overrightarrow{l_i}=(\text{0}\text{.10 m) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0 m) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0 m) <mover accent="true"><mo>k</mo><mo>^</mo></mover>}$

Length of the wire after rotating $=\overrightarrow{l_f}=(0\text{ m) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0}\text{.10 m) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0 m) <mover accent="true"><mo>k</mo><mo>^</mo></mover>}$

Magnetic field $=\overrightarrow{B}=a\widehat{i}+b\widehat{j}+c\widehat{k}$

Magnetic force before rotating $=\overrightarrow{F_i}=(0\text{ N})\widehat{i}+(3\text{ N})\widehat{i}+(2\text{ N})\widehat{k}$

Magnetic force after rotating $=\overrightarrow{F_f}=(0\text{ N})\widehat{i}-(3\text{ N})\widehat{j}-(2\text{ N})\widehat{k}$

Formula Used:

Magnetic force on a current carrying wire is given as

  $\overrightarrow{F}=I(\overrightarrow{l}\times \overrightarrow{B})$

Calculation:

Magnetic force before rotating is given as

  $\begin{array}{l}\overrightarrow{F_i}=I(\overrightarrow{l_i}\times \overrightarrow{B})\\ (0)\widehat{i}+(3)\widehat{j}+(2)\widehat{k}=(2)(((\text{0}\text{.10 ) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0 ) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0 ) <mover accent="true"><mo>k</mo><mo>^</mo></mover>})\times (a\widehat{i}+b\widehat{j}+c\widehat{k}))\\ (0)\widehat{i}+(1.5)\widehat{j}+(1)\widehat{k}=(((\text{0}\text{.10 ) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0 ) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0 ) <mover accent="true"><mo>k</mo><mo>^</mo></mover>})\times (a\widehat{i}+b\widehat{j}+c\widehat{k}))\\ (1.5)\widehat{j}+(1)\widehat{k}=(\text{0}\text{.10 )}b\widehat{k}-(\text{0}\text{.10 )}c\widehat{j}\\ \text{ comparing both side }\\ (\text{0}\text{.10 )}b=1\text{ and }-(\text{0}\text{.10 )}c=1.5\\ b=10\text{ and c = -15}\end{array}$

Magnetic force after rotating is given as

  $\begin{array}{l}\overrightarrow{F_f}=I(\overrightarrow{l_f}\times \overrightarrow{B})\\ (0)\widehat{i}-(3)\widehat{j}-(2)\widehat{k}=(2)\left[\left((0\text{) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0}\text{.10) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0) <mover accent="true"><mo>k</mo><mo>^</mo></mover>}\right)\times (a\widehat{i}+b\widehat{j}+c\widehat{k})\right]\\ (0)\widehat{i}-(1.5)\widehat{j}-(1)\widehat{k}=\left[(0\text{) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0}\text{.10) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0) <mover accent="true"><mo>k</mo><mo>^</mo></mover>}\right]\times (a\widehat{i}+10\widehat{j}-15\widehat{k})\\ -(1.5)\widehat{j}-(1)\widehat{k}=(-(0.10)a\widehat{k})-(1.5)\widehat{i}\\ \text{ comparing both side }\\ -(0.10)a=-1\\ \text{a}\text{=10}\end{array}$

Hence the magnetic field is given as

  $\begin{array}{l}\overrightarrow{B}=a\widehat{i}+b\widehat{j}+c\widehat{k}\\ \overrightarrow{B}=\left(10\widehat{i}+10\widehat{j}-15\widehat{k}\right)\text{T}\end{array}$

Conclusion:

The magnetic field comes out to be $\left(10\widehat{i}+10\widehat{j}-15\widehat{k}\right)\text{T}$ .