The magnetic field is a "pseudovector" (more properly, a 2-form), as opposed to the electric field, which is a vector (or a 1-form). That is, under parity, $\mathbf B$ is left unchanged. You can see this from the Lorentz force, $$\mathbf F = q(\mathbf E + \mathbf v\times \mathbf B)$$ where since force is a vector, $\mathbf E$ must also be a vector. Since $\mathbf v \times \mathbf B$ is a product, if $\mathbf B$ was a vector, this term would not transform properly under parity. Thus $\mathbf B$ does not change when we perform a parity transformation.

However I think the more correct way to see this is from the relativistic formulation of electrodynamics. Introduce the electromagnetic field tensor $$F_{\mu\nu} =\begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix}$$ and note that the purely spatial part $F_{ij}$, $1 \le i, j \le 3$ is equivalent to the magnetic field. Since there are two indices, the components are invariant under parity transformations. The electric field is given by $E_i = F_{0i}$, so it changes sign under parity, it is a vector.

The more sophisticated yet way to see this decomposition is that if there is timelike 1-form $dt$ we can decompose the field strength 2-form as $$F = E\wedge dt + B.$$ We see that $E$ is a 1-form (equivalent to a vector after raising the index), but $B$ is a 2-form (often called pseudovector, because not enough people know about the wonders of differential forms).

Now your transformation is not quite $P : (x,y,z) \mapsto (-x,-y,-z).$ It is $RP : (x,y,z) \mapsto (-x,y,z)$, or $P$, then a rotation by $\pi$ in the $yz$-plane. Here the $x$-axis is along $\mathbf v$ and the $z$-axis along $\mathbf B$. Since $\mathbf v$ is perpendicular to the plane of the rotation, it is just affected by the reflection, and is to the left. $\mathbf B$, lying in the plane of rotation, is rotated half a revolution, and is now out of the page, so the force is upward. The force being a vector, this is precisely what we expect. I hope this answers your question.

The magnetic field is a "pseudovector" (more properly, a 2-form), as opposed to the electric field, which is a vector (or a 1-form). That is, under parity, $\mathbf B$ is left unchanged. You can see this from the Lorentz force, $$\mathbf F = q(\mathbf E + \mathbf v\times \mathbf B)$$ where since force is a vector, $\mathbf E$ must also be a vector. Since $\mathbf v \times \mathbf B$ is a product, if $\mathbf B$ were a vector, this term would not transform properly under parity. Thus $\mathbf B$ does not change when we perform a parity transformation.

However I think the more correct way to see this is from the relativistic formulation of electrodynamics. Introduce the electromagnetic field tensor $$F_{\mu\nu} =\begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix}$$ and note that the purely spatial part $F_{ij}$, $1 \le i, j \le 3$ is equivalent to the magnetic field. Since there are two indices, the components are invariant under parity transformations. The electric field is given by $E_i = F_{0i}$, so it changes sign under parity, it is a vector.

The more sophisticated yet way to see this decomposition is that if there is timelike 1-form $dt$ we can decompose the field strength 2-form as $$F = E\wedge dt + B.$$ We see that $E$ is a 1-form (equivalent to a vector after raising the index), but $B$ is a 2-form (often called pseudovector, because not enough people know about the wonders of differential forms).

Now your transformation is not quite $P : (x,y,z) \mapsto (-x,-y,-z).$ It is $RP : (x,y,z) \mapsto (-x,y,z)$, or $P$, then a rotation by $\pi$ in the $yz$-plane. Here the $x$-axis is along $\mathbf v$ and the $z$-axis along $\mathbf B$. Since $\mathbf v$ is perpendicular to the plane of the rotation, it is just affected by the reflection, and is to the left. $\mathbf B$, lying in the plane of rotation, is rotated half a revolution, and is now out of the page, so the force is upward. The force being a vector, this is precisely what we expect. I hope this answers your question.

The magnetic field is a "pseudovector" (more properly, a 2-form), as opposed to the electric field, which is a vector (or a 1-form). That is, under parity, $\mathbf B$ is left unchanged. You can see this from the Lorentz force, $$\mathbf F = q(\mathbf E + \mathbf v\times \mathbf B)$$ where since force is a vector, $\mathbf E$ must also be a vector. Since $\mathbf v \times \mathbf B$ is a product, if $\mathbf B$ was a vector, this term would not transform properly under parity. Thus $\mathbf B$ does not change when we perform a parity transformation.

However I think the more correct way to see this is from the relativistic formulation of electrodynamics. Introduce the electromagnetic field tensor $$F_{\mu\nu} =\begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix}$$ and note that the purely spatial part $F_{ij}$, $1 \le i, j \le 3$ is equivalent to the magnetic field. Since there are two indices, the components are invariant under parity transformations. The electric field is given by $E_i = F_{0i}$, so it changes sign under parity, it is a vector.

The more sophisticated yet way to see this decomposition is that if there is timelike 1-form $dt$ we can decompose the field strength 2-form as $$F = E\wedge dt + B.$$ We see that $E$ is a 1-form (equivalent to a vector after raising the index), but $B$ is a 2-form (often called pseudovector, because not enough people know about the wonders of differential forms).

The magnetic field is a "pseudovector" (more properly, a 2-form), as opposed to the electric field, which is a vector (or a 1-form). That is, under parity, $\mathbf B$ is left unchanged. You can see this from the Lorentz force, $$\mathbf F = q(\mathbf E + \mathbf v\times \mathbf B)$$ where since force is a vector, $\mathbf E$ must also be a vector. Since $\mathbf v \times \mathbf B$ is a product, if $\mathbf B$ was a vector, this term would not transform properly under parity. Thus $\mathbf B$ does not change when we perform a parity transformation.

However I think the more correct way to see this is from the relativistic formulation of electrodynamics. Introduce the electromagnetic field tensor $$F_{\mu\nu} =\begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix}$$ and note that the purely spatial part $F_{ij}$, $1 \le i, j \le 3$ is equivalent to the magnetic field. Since there are two indices, the components are invariant under parity transformations. The electric field is given by $E_i = F_{0i}$, so it changes sign under parity, it is a vector.

The more sophisticated yet way to see this decomposition is that if there is timelike 1-form $dt$ we can decompose the field strength 2-form as $$F = E\wedge dt + B.$$ We see that $E$ is a 1-form (equivalent to a vector after raising the index), but $B$ is a 2-form (often called pseudovector, because not enough people know about the wonders of differential forms).

Now your transformation is not quite $P : (x,y,z) \mapsto (-x,-y,-z).$ It is $RP : (x,y,z) \mapsto (-x,y,z)$, or $P$, then a rotation by $\pi$ in the $yz$-plane. Here the $x$-axis is along $\mathbf v$ and the $z$-axis along $\mathbf B$. Since $\mathbf v$ is perpendicular to the plane of the rotation, it is just affected by the reflection, and is to the left. $\mathbf B$, lying in the plane of rotation, is rotated half a revolution, and is now out of the page, so the force is upward. The force being a vector, this is precisely what we expect. I hope this answers your question.