A wire carrying current has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius is lying in plane. Magnetic field at point O is [2015]

Question

A wire carrying current $I$ has the shape as shown in adjoining figure.
Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius $R$ is lying in $Y - Z$ plane. Magnetic field at point O is [2015]

Smart Achievers · Accepted Answer

(1)

Given situation is shown in the figure.

Parallel wires 1 and 3 are semi-infinite, so magnetic field at O due to them

${\vec{B}}_{1} = {\vec{B}}_{3} = - \frac{μ_{0} I}{4 π R} \hat{k}$

Magnetic field at O due to semi-circular arc in YZ-plane is given by ${\vec{B}}_{2} = - \frac{μ_{0} I}{4 R} \hat{i}$

Net magnetic field at point O is given by

$\vec{B} = {\vec{B}}_{1} + {\vec{B}}_{2} + {\vec{B}}_{3}$

$= - \frac{μ_{0} I}{4 π R} \hat{k} - \frac{μ_{0} I}{4 R} \hat{i} - \frac{μ_{0} I}{4 π R} \hat{k} = - \frac{μ_{0} I}{4 π R} (π \hat{i} + 2 \hat{k})$

Solution

Q.
A wire carrying current $I$ has the shape as shown in adjoining figure.
Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius $R$ is lying in $Y - Z$ plane. Magnetic field at point O is [2015]

1 $\vec{B}$ $= - \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} + 2 \hat{k})$

2 $\vec{B}$ $= \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} - 2 \hat{k})$

3 $\vec{B}$ $= \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} + 2 \hat{k})$

4 $\vec{B}$ $= - \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} - 2 \hat{k})$

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Solution

Q. A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is [2015]

1 B→=-μ04πIR(πi^+2k^)

2 B→=μ04πIR(πi^-2k^)

3 B→=μ04πIR(πi^+2k^)