91Ó°ÊÓ

a) The current flowing through the conductor is $I=-5.0\text{A}$

b) The magnetic field is $\stackrel{→}{B}=\left(3.0\hat{i}+8.0x^2\hat{j}\text{mT}\right)\text{or}\left(3.0\hat{i}+8.0x^2\hat{j}\right)×{10}^{-3}\text{T}$

c) The segment of the conductor on which force is acting as $l=2.0\text{m}$

d) The conductor lies in between is $x_i=1.0\text{m}$ and $x_f=3.0\text{m}$

If a particle is moving with a uniform velocity within a uniform magnetic field, then it experiences a magnetic force due to its charge value that induces a current within the loop. The force acting on the particle is due to the current along the length of the conductive wire. The direction of this magnetic force is given by Fleming's right-hand rule as the force is perpendicular to both the speed and magnetic field acting on it.

The magnetic force along a loop wire is as follows:

$\stackrel{→}{d{F}_B}=i\left(\stackrel{→}{dL}×\stackrel{→}{B}\right)$ …… (i)

Here, $i$is the current in the loop, $\stackrel{→}{B}$is the magnetic field vector that it experiences, $\stackrel{→}{dL}$ is the length vector of the conducting wire.

The conductor is lying along the x axis. Hence,$\stackrel{→}{dL}=dx\hat{i}\text{m}$and$\stackrel{→}{B}=\left(B_x\hat{i}+B_y\hat{j}\right)\text{T}$

The expression of the magnetic force on the current carrying wire is given using the above data in equation (i) as follows:

$\stackrel{→}{F_B}=∫i\left(dx\hat{i}\text{m}\right)×\left(\left(B_x\hat{i}+B_y\hat{j}\right)\text{T}\right)$

According to the property of the vector product of two vectors,$\hat{i}×\hat{i}=0$and$\hat{i}×\hat{j}=\hat{k}$. The force value can be given using the given data in the above equation as follows:

$\stackrel{→}{F_B}=$$i{∫}_{xi}^{xf}{B}_{Y}dx\hat{k}$

role="math" localid="1662526547252" $=\left(-5.0A\right){∫}_{1.0}^{3.0}\left[\left(\left(8.0×{10}^{-3}{x}^{2}T\right)\left(dxm\right)\right)\right]\hat{k}$

$=\left[\left(-5.0\right)×\left(8.0\right)×{\left(\frac{x^3}{3}\right)}_{1.0}^{3.0}\hat{k}×{10}^{-3}\right]N$

$=\left[\left(-5.0\right)×\left(8.0\right)××{10}^{-3}\left[\frac{{\left(3.0\right)}^3}{3}-\frac{{\left(1.0\right)}^3}{3}\right]\hat{k}\right]N$

Substitute the values and solve as:

$$\begin{gathered} \stackrel{→}{F_B}=\left[-40××{10}^{-3}\left(\frac{27-1}{3}\right)\hat{k}\right]\text{N} \\ =\left(-0.346\hat{k}\right)\text{N} \\ ≈\left(-0.35\hat{k}\right)\text{N} \end{gathered}$$

.

Hence, the value of the force is $\left(-0.35\hat{k}\right)\text{N}$.