Q27E - University Physics with Modern Physics

Question

Two very long insulated wires perpendicular to each other in the same plane carry currents as shown in Fig. Find the magnitude of the net magnetic field these wires produce at points P and Q if the 10.0A current  to the left.

Step-by-Step Solution

Answer

The net magnetic field P and Q is $9.0\times {10}^{-6}T$ .

Step 1: The magnetic field due to wire at a point

The magnetic field due to wire at point is given by

 

$\mathrm{B}=\frac{{\mu }_0\mathrm{l}}{2\pi }$                   

 

Where, B is the magnetic field due to wire, ${\mu }_0$ is the permeability of vaccum, l is the current through the wire and r is the distance from wire.

 

Sign convention for magnetic field

 

 Conventionally for the direction of inward and outward magnetic field to the plane negative or positive sign is considered for magnetic field.

 

 

Magnetic field due to wire 1  is given by

                                       

${\mathrm{B}}_1=-\frac{{\mu }_0{I}_1}{2\pi r}$

_{Here, Magnetic field is negative due to inward direction of magnetic field with plane at both points P and Q.}

 

Magnetic field due to wire 2  at is given by

 

Given : The current flow in wire 1 is $l_1=12 A$ . 

             The current flow in wire 2 is $l_2=10A$ _. 

                        

${\mathrm{B}}_2=\frac{{\mu }_0{\mathrm{l}}_2}{2\pi }$

_{Here, Magnetic field is positive due to outward direction of magnetic field with plane at both points P and Q.}

Step 2: Calculation of magnetic field at point P and Q

The magnetic field at point P is given by

 

$\begin{array}{r}{B}_p=B_2-B_1 \\ B_p=\frac{{\mu }_0}{2\pi }\left(\frac{l_2}{r_2}-\frac{l_1}{r_1}\right)\end{array}$ 

Putting the values of constants in above equation 

 

$\begin{array}{r}{B}_p=\frac{4\pi \times {10}^{-7}}{2\pi }\left(\frac{10}{0.080}-\frac{12}{0.15}\right)\\ B_p=9.0\times {10}^{-6}\mathrm{T} \end{array}$

 

Thus , the magnetic field at point p is $9.0\times {10}^{-6}T$ 

 

The magnetic field at point Q is given by

$\begin{array}{r}{B}_Q=B_2-B_1 \\ B_Q=\frac{{\mu }_0}{2\pi }\left(\frac{I_2}{r_2}-\frac{l_1}{r_1}\right)\end{array}$ 

 

Putting the values of constants in above equation 

 

$\begin{array}{r}{B}_Q=\frac{4\pi \times {10}^{-7}}{2\pi }\left(\frac{10}{0.080}-\frac{12}{0.15}\right)\\ B_Q=9.0\times {10}^{-6}\mathrm{T}\end{array}$

 

Thus , the magnetic field at point Q is $9.0\times {10}^{-6}T$ .