Question

Please answer this question.

Answer 1

Topic :-

Magnetic Effect of Current

Given :-

A figure.

To Find :-

Magnetic Field at the point O.

Formulae :-

Magnetic Field due to Current Carrying Straight Wire

$B=\dfrac{ \mu_0i(sin \alpha+ sin \beta)}{4 \pi r}$

where

$\alpha$ and $\beta$ are angles to the extremities of wire measured perpendicularly from wire.

r is perpendicular distance of wire from point.

i is current in wire.

Here in figure, we can see that

$\alpha$ and $\beta$ are $90^{\circ}\:and\:0^{\circ}$ respectively.

So, formula for above case of wire will be

$B=\dfrac{ \mu_0i(sin90^{\circ}+ sin 0^{\circ})}{4 \pi r}$

$B=\dfrac{ \mu_0i(1+ 0)}{4 \pi r}$

$B=\dfrac{ \mu_0i}{4 \pi r}$

Magnetic Filed due to Current Carrying Semicircular Wire

$B=\dfrac{ \mu_0i}{4r}$

where,

r is radius of semicircle.

i is current in wire.

Note :- Direction of Magnetic Field will be calculated with the help of Right Hand Thumb Rule.

Solution :-

With the help of superposition, Magnetic Field at point 'O' will be

Magnetic Field Due to Straight Wire 1 + Magnetic Field Due to semicircular wire + Magnetic Field Due to Straight Wire 2

Here,

Radius of Semicircle = r

Distance of point from wire = r

Current in wire = I

Applying the formula,

$\dfrac{ \mu_0I}{4 \pi r}(\hat{k})+\dfrac{ \mu_0I}{4r}(\hat{k})+\dfrac{ \mu_0I}{4 \pi r}(\hat{k})$

$\dfrac{ \mu_0I}{2 \pi r}(\hat{k})+\dfrac{ \mu_0I}{4r}(\hat{k})$

$\dfrac{ \mu_0I}{2r} \left ( \dfrac{1}{ \pi}+\dfrac{1}{2} \right )(\hat{k})$

$\dfrac{ \mu_0I(2+ \pi)}{4 \pi r}(\hat{k})$

Answer :-

So, magnetic field at point O is

$\dfrac{ \mu_0I(2+ \pi)}{4 \pi r}(\hat{k})$

Note :- X - Y plane is taken in plane of screen.