Name: Magnetomotive Force
Uploaded: 2019-07-19
Duration: 2 min 40 s
Description: You might have seen the equation of magnetic flux. There appears a quantity which is known as magnetomotive force. Let's watch the video to learn more.

Question

Find magnetic field at point $P$ in following cases.

Accepted Answer

According to the question the given figures are:-

Now, from above figures we need to find the value of magnetic field at point $P .$

First designate the given current point $2$ and $3 .$

Now,

$\Rightarrow \vec{B_{2}} = - \vec{B_{3}}$

Apply bio-Savart law.

$\Rightarrow {\vec{B}}_{n e t} = \vec{B_{1}} + \vec{B_{2}} + \vec{B_{3}} \Rightarrow {\vec{B}}_{n e t} = \frac{μ_{o} i}{4 r} (\hat{k}) + \frac{μ_{o} i}{4 πR} (\hat{k}) + \frac{μ_{o} i}{4 πR} (\hat{- k}) \Rightarrow {\vec{B}}_{n e t} = \frac{μ_{o} i}{4 r} (\hat{k}) .$

$\hat{k}$ is showing out of the page.

Magnetic field for a circle $\vec{B =} \frac{μ_{0} i}{4} \frac{2 i}{r} .$

For semicircle $= \frac{μ_{0}}{4} \frac{2 i}{r} \times \frac{1}{2} = \frac{μ_{0} i}{4 r} .$

Where $r =$ Radius of the circle.

$i =$ Current

$\Rightarrow \vec{B} \propto \frac{1}{r} .$

$\Rightarrow R_{1} < R_{2} \Rightarrow B_{1} > B_{2}$

Now, $\Rightarrow B_{1} = \frac{μ_{0} i}{4 (R_{1})} . \Rightarrow B_{2} = \frac{μ_{0} i}{4 (R_{2})} \Rightarrow B_{1} - B_{2} = \frac{μ_{0} i}{4 (R_{1})} - \frac{μ_{0} i}{4 (R_{2})} = \frac{μ_{0} i}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}}) .$

Outwards.

For SR & QP the field are equal and oppositie. So they will cancel to each other.

$(a)$ $\frac{μ_{0} i}{8 r}$ into the page

$(b)$ $\frac{3 μ_{0} i}{8 r}$ into the page

$(c)$ $\frac{μ_{0} I}{2 π r} + \frac{μ_{0} i}{4 R}$ outside the page

$(d)$ $\frac{μ_{0} i}{4 R}$ out of the page

$(e)$ Zero

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