A long straight wire along the Z-axis carries a current in the negative Z-direction. The magnetic vector field at a point having coordinates in the plane is [2002]

Question

A long straight wire along the Z-axis carries a current $I$ in the negative Z-direction. The magnetic vector field $\vec{B}$ at a point having coordinates $(x, y)$ in the $Z = 0$ plane is [2002]

Smart Achievers · Accepted Answer

(1)

The wire carries a current $\vec{B}$ in the negative $z$ -direction. We have to consider the magnetic vector field $\vec{B}$ at $(x, y)$ in the $z = 0$ plane.

Magnetic field $\vec{B}$ is perpendicular to $O P$ .

$∴$ $\vec{B} = B \sin θ \hat{i} - B \cos θ \hat{j}$

$\sin θ = \frac{y}{r}, \cos θ = \frac{x}{r}, B = \frac{μ_{0} I}{2 π r}, x^{2} + y^{2} = r^{2}$

$∴$ $\vec{B} = \frac{μ_{0} I}{2 π r^{2}} (y \hat{i} - x \hat{j})$

or $\vec{B} = \frac{μ_{0} I (y \hat{i} - x \hat{j})}{2 π (x^{2} + y^{2})}$

Solution

Q.
A long straight wire along the Z-axis carries a current $I$ in the negative Z-direction. The magnetic vector field $\vec{B}$ at a point having coordinates $(x, y)$ in the $Z = 0$ plane is [2002]

1 $\frac{μ_{0} I}{2 π} (\frac{y \hat{i} - x \hat{j}}{x^{2} + y^{2}})$

2 $\frac{μ_{0} I}{2 π} (\frac{x \hat{i} + y \hat{j}}{x^{2} + y^{2}})$

3 $\frac{μ_{0} I}{2 π} (\frac{x \hat{j} - y \hat{i}}{x^{2} + y^{2}})$

4 $\frac{μ_{0} I}{2 π} (\frac{x \hat{i} - y \hat{j}}{x^{2} + y^{2}})$

Want Us to Email you About Special Offers & Updates?

Solution

Q. A long straight wire along the Z-axis carries a current I in the negative Z-direction. The magnetic vector field B→ at a point having coordinates (x,y) in the Z=0 plane is [2002]

1 μ0I2π(yi^-xj^x2+y2)

2 μ0I2π(xi^+yj^x2+y2)

3 μ0I2π(xj^-yi^x2+y2)