An infinitely long wire, located on the -axis, carries a current along the -direction and produces the magnetic field . The magnitude of the line integral along a straight line from the point to is given by [ is the magnetic permeability of free space.

Question

An infinitely long wire, located on the $z$ -axis, carries a current $I$ along the $+ z$ -direction and produces the magnetic field $\vec{B}$ . The magnitude of the line integral $\int \vec{B} \cdot d \vec{l}$ along a straight line from the point $(\sqrt{3} a, a, 0)$ to $(a, a, 0)$ is given by [ $μ_{0}$ is the magnetic permeability of free space.] [2024]

Smart Achievers · Accepted Answer

(1)

From figure,

$| \vec{d ℓ} |$ $= r d θ$

Magnetic field, $| \vec{B} |$ $= \frac{μ_{0} I}{2 π r}$

$\int \vec{B} \cdot d \vec{ℓ} = \int | \vec{B} | | \vec{d ℓ} | \cos θ ° = \int (\frac{μ_{0} I}{2 π r}) (r d θ)$

$= \int_{θ_{1}}^{θ_{2}} \frac{μ_{0} I}{2 π} d θ = \frac{μ_{0} I}{2 π} [θ_{2} - (- θ_{1})]$ $[θ_{1} is anticlockwise hence taken negative]$

or, $\int \vec{B} \cdot d \vec{ℓ} = \frac{μ_{0} I}{2 π} (θ_{2} + θ_{1})$

Here, $\tan θ_{1} = \frac{a \sqrt{3}}{a} = \sqrt{3}$ $∴$ $θ_{1} = \frac{π}{3}$

and $\tan θ_{2} = \frac{a}{a} = 1$ $∴$ $θ_{2} = \frac{π}{4}$

Putting the values of $θ_{1}$ and $θ_{2}$ in Eq. (i),

$\int \vec{B} \cdot d \vec{ℓ} = \frac{μ_{0} I}{2 π} [\frac{π}{3} + \frac{π}{4}]$

or, $\int \vec{B} \cdot \vec{d ℓ} = \frac{μ_{0} I}{2 π} [\frac{4 π + 3 π}{12}] = \frac{7 μ_{0} I}{24}$

Solution

1 $\frac{7 μ_{0} I}{24}$

2 $\frac{7 μ_{0} I}{12}$

3 $\frac{μ_{0} I}{8}$

4 $\frac{μ_{0} I}{6}$

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Solution

1 7μ0I24

2 7μ0I12

3 μ0I8

4 μ0I6

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1 $\frac{7 μ_{0} I}{24}$

2 $\frac{7 μ_{0} I}{12}$

3 $\frac{μ_{0} I}{8}$

4 $\frac{μ_{0} I}{6}$