Chapter 26, Problem 24P

To determine

The magnetic field.

Answer to Problem 24P

  $\left(0.5\widehat{i}+0.5\widehat{j}+0\widehat{k}\right)\text{T}$

Explanation of Solution

Given:

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

Magnitude of current in the wire $=I=4\text{ A}$

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

Length of the wire after rotating $=\overrightarrow{l_f}=(0.10\text{ m) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0 m) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0 m) <mover accent="true"><mo>k</mo><mo>^</mo></mover>}$

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

Magnetic force before rotating $=\overrightarrow{F_i}=-(0.20\text{ N})\widehat{i}+(0.20\text{ N})\widehat{i}+(0\text{ N})\widehat{k}$

Magnetic force after rotating $=\overrightarrow{F_f}=(0\text{ N})\widehat{i}+(0\text{ N})\widehat{j}+(0.20\text{ N})\widehat{k}$

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

  $\begin{array}{l}\overrightarrow{F_i}=I(\overrightarrow{l_i}\times \overrightarrow{B})\\ -(0.20\text{ N})\widehat{i}+(0.20\text{ N})\text{<mover accent="true"><mo>j</mo><mo>^</mo></mover>}+(0\text{ N})\widehat{k}=(4)\left[\left((\text{0) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0}\text{.10) <mover accent="true"><mo>k</mo><mo>^</mo></mover>}\right)\times (a\widehat{i}+b\widehat{j}+c\widehat{k})\right]\\ -(0.05\text{ N})\widehat{i}+(0.05\text{ N})\text{<mover accent="true"><mo>j</mo><mo>^</mo></mover>}+(0\text{ N})\widehat{k}=\left[(\left(\text{0) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0}\text{.10) <mover accent="true"><mo>k</mo><mo>^</mo></mover>}\right)\times (a\widehat{i}+b\widehat{j}+c\widehat{k})\right]\\ -(0.05\text{ N})\widehat{i}+(0.05\text{ N})\text{<mover accent="true"><mo>j</mo><mo>^</mo></mover>}=(\text{0}\text{.10)}a\widehat{j}-(\text{0}\text{.10)b}\widehat{i}\\ \text{ Comparing coefficient of }\widehat{i}\text{ and }\widehat{j}\text{ both side }\\ -(\text{0}\text{.10 )}b=-0.05\text{ and }(\text{0}\text{.10 )a}=0.05\\ b=0.5\text{ and a = 0}\text{.5}\end{array}$

Magnetic force after rotating is given as

  $\begin{array}{l}\overrightarrow{F_f}=I(\overrightarrow{l_f}\times \overrightarrow{B})\\ (0)\widehat{i}+(0)\widehat{j}+(0.20)\widehat{k}=(4)\left[\left((0.10\text{) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0) <mover accent="true"><mo>k</mo><mo>^</mo></mover>}\right)\times (a\widehat{i}+b\widehat{j}+c\widehat{k})\right]\\ (0)\widehat{i}+(0)\widehat{j}+(0.05)\widehat{k}=\left[\left((0.10\text{) <mover accent="true"><mo>i</mo><mo>^</mo></mover><mo> + </mo>}(\text{0) <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> + </mo>}(\text{0) <mover accent="true"><mo>k</mo><mo>^</mo></mover>}\right)\times (a\widehat{i}+b\widehat{j}+c\widehat{k})\right]\\ (0)\widehat{i}+(0)\widehat{j}+(0.05)\widehat{k}=(0.10)b\widehat{k}-(1.5)c\text{<mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> </mo>}\\ \text{ Comparing coefficient of <mover accent="true"><mo>j</mo><mo>^</mo></mover><mo> both side </mo>}\\ -(1.5)c=0\\ \text{c = 0}\end{array}$

Hence the magnetic field is given as

  $\begin{array}{l}\overrightarrow{B}=a\widehat{i}+b\widehat{j}+c\widehat{k}\\ \overrightarrow{B}=\left(0.5\widehat{i}+0.5\widehat{j}+0\widehat{k}\right)\text{T}\end{array}$

Conclusion:

The magnetic field comes out to be $\left(0.5\widehat{i}+0.5\widehat{j}+0\widehat{k}\right)\text{T}$ .