91Ó°ÊÓ

a) The length of wire, $L=50.0\text{cm or}50.0×{10}^{-2}\text{m}$

b) The current flowing through the wire, $i=0.500\text{A}$

c) The magnetic field vector, $\stackrel{→}{B}=\left(0.00300\text{T}\right)\hat{j}+\left(0.0100\text{T}\right)\hat{k}$

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 wire is kept in the x-axis, where the current is flowing in the positive x-direction.

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

$\stackrel{→}{F_B}=iL\hat{i}×\left(B_y\hat{j}+B_z\hat{k}\right)$ ……………………………. (ii)

According to the vector product of two vectors, obtain the magnetic field as:

$\hat{i}×\left(B_y\hat{j}+B_z\hat{k}\right)=-B_z\hat{j}+B_y\hat{k}$

Thus, the force equation (ii) now can be given using the given data as follows:

$$\begin{gathered} \stackrel{→}{F_B}=iL\left(-B_z\hat{j}+B_y\hat{k}\right) \\ =\left(0.500\text{A}×50.0×{10}^{-2}\text{m}\right)\left[-\left(0.0100\text{T}\right)\hat{j}+\left(0.00300\text{T}\right)\hat{k}\right] \\ =\left[\left(-250×{10}^{-3}\right)\hat{j}+\left(0.750×{10}^{-3}\right)\hat{k}\right]\text{N} \end{gathered}$$

Hence, the value of the magnitude of the force is $\left[\left(-250×{10}^{-3}\right)\hat{j}+\left(0.750×{10}^{-3}\right)\hat{k}\right]\text{N}$.