The magnetic field is uniform over each face of the box shown in $FIGUREEX31.6.$ What are the magnetic field strength and direction on the front surface?

We have been given that the magnetic field is uniform over each face of the box shown in $FIGUREEX31.6$.

We need to find the magnetic field strength and direction on the front surface.

We know the formula: $\oint B.dA=0$ (Let this equation be ($1$))

For direction, we know that by convention, the flux coming in the surface is negative and the flux going out of the surface is positive.

Apply the formula given in equation $\left(1\right)$on all the surfaces of the box:

Left surface: localid="1650206016144" role="math" $3T(2\times {10}^{-6}{m}^2)=6\times {10}^{-6}T{m}^2$

Right surface: localid="1650205102517" $3T(2\times {10}^{-6}{m}^2)=6\times {10}^{-6}T{m}^2$

Top surface: localid="1650205113372" $-2T(2\times {10}^{-6}{m}^2)=-4\times {10}^{-6}T{m}^2$

Bottom surface: localid="1650205124248" $-1T(2\times {10}^{-6}{m}^2)=-2\times {10}^{-6}T{m}^2$

Back surface: localid="1650205133312" $-2T(1\times {10}^{-6}{m}^2)=-2\times {10}^{-6}T{m}^2$

Let magnetic field strength on the front surface be $B_f$.

Front surface: $B_{f}T(1\times {10}^{-6}{m}^2)=B_f\times {10}^{-6}T{m}^2$

The net flux through all the surfaces of the box should be zero. So we can calculate the magnetic field strength on the front surface $\left(B_f\right)$by equating the sum of flux of all the surfaces to zero.

$(6T+6T-4T-2T-2T+B_{f}T)({10}^{-6}{m}^2)=0$

Dividing ${10}^{-6}{m}^2$on both sides, we get:

$12T-8T+B_f=0$

$4T+B_f=0$

$B_f=-4T$

The magnetic field strength on the front surface is $B_f=-4T$.

Since it is negative, it means that the magnetic field is coming into the front surface.