Respuesta :
Answer:
The magnitude of net magnetic field is [tex]3.85\times10^{-5}\ T[/tex]
Explanation:
Given that,
Current = 23.0 A
Distance = 23.0 cm
Angle = 46°
Magnetic field [tex]B= 5.0\times10^{-5}\ T[/tex]
We need to calculate the magnetic field due to the wire
Using formula of magnetic field
[tex]B=\dfrac{\mu_{0}I}{2\pi r}[/tex]
Put the value into the formula
[tex]B_{w}=\dfrac{4\pi\times10^{-7}\times23.0}{2\pi\times23.0\times10^{-2}}[/tex]
[tex]B_{w}=0.00002\ T[/tex]
[tex]B_{w}=2.0\times10^{-5}\ T[/tex]
We need to calculate the horizontal component of the magnetic field
Using formula of magnetic field
[tex]B_{net}x=B_{e}\cos\theta[/tex]
Put the value into the formula
[tex]B_{net}x=5.0\times10^{-5}\times\cos45[/tex]
[tex]B_{net}x=0.0000353\ T[/tex]
[tex]B_{net}x=3.53\times10^{-5}\ T[/tex]
We need to calculate the component of the magnetic field
Using formula of magnetic field
[tex]B_{net}y=B_{w}-B\sin\theta[/tex]
Put the value into the formula
[tex]B_{net}y=2.0\times10^{-5}-(5.0\times10^{-5})\times\sin45[/tex]
[tex]B_{net}y=-0.0000153\ T[/tex]
[tex]B_{net}y=-1.53\times10^{-5}\ T[/tex]
We need to calculate the magnitude of net magnetic field
Using formula of magnetic field
[tex]B_{net}=\sqrt{(B_{net}x)^2+(B_{net}y)^2}[/tex]
Put the value into the formula
[tex]B_{net}=\sqrt{(3.53\times10^{-5})^2+(-1.53\times10^{-5})^2}[/tex]
[tex]B_{net}=0.0000385\ T[/tex]
[tex]B_{net}=3.85\times10^{-5}\ T[/tex]
Hence, The magnitude of net magnetic field is [tex]3.85\times10^{-5}\ T[/tex]
The magnitude of the net magnetic field 23 cm due west of the wire is obtained to be [tex]3.81\times 10^{-5}\,T[/tex].
Biot-Savart Law
According to Biot-Savart law, the magnetic field due to current in a straight wire is given by;
[tex]B=\frac{\mu _0 I}{2 \pi r}[/tex]
Substituting the given values;
[tex]B_W = \frac{4\pi \times 10^{-7}\times 23}{2\pi \times 23\times 10^{-2}} =2\times10^{-5}\,T[/tex]
The x-component of earth's magnetic field at the given point;
[tex]B_x= B_H cos \theta = 5\times 10^{-5}\times cos \, 46^\circ=3.47\times 10^{-5}\,T[/tex]
The y-component of earth's magnetic field at the given point;
[tex]B_y= B_W-B_H \,sin \theta = (2\times 10^{-5})-(5\times 10^{-5}\times sin \, 46^\circ)=-1.59\times 10^{-5}\,T[/tex]
So the magnitude of the net magnetic field is given by;
[tex]B_{net}=\sqrt{(3.47\times 10^{-5})^2 + (-1.59\times 10^{-5})^2\,} = 3.81\times 10^{-5}\,T[/tex]
Learn more about Biot-Savart law here:
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