Question

A straight wire of finite length carrying current i substance on angle of 60 degree at point p as shown the magnetic field at p is

Answer 1

Answer:

The magnetic field at p is $\dfrac{\mu_{0}I}{2\sqrt{3}x}$.

Explanation:

Given that,

Angle = 60°

Current = I

Suppose the figure is,

According to figure,

$\dfrac{r}{x}=\cos30^{\circ}$

$r=x\dfrac{\sqrt{3}}{2}$

We need to calculate the magnetic field

Using formula of magnetic field

$\vec{B}=\dfrac{\mu_{0}I}{4\pi r}(\sin\theta_{1}+\sin\theta_{2})$

Put the angle into the formula

$\vec{B}=\dfrac{\mu_{0}I}{4\pi\times x\dfrac{\sqrt{3}}{2}}(\sin30+\sin30)$

$\vec{B}=\dfrac{\mu_{0}I}{2\sqrt{3}x}\times(2\sin30)$

$\vec{B}=\dfrac{\mu_{0}I}{2\sqrt{3}x}$

Hence, The magnetic field at p is $\dfrac{\mu_{0}I}{2\sqrt{3}x}$.