Question

A conductor carrying a current I is oftypes as shown below. Find themagnetic field induction at the commoncentre o of all the three arcs.​

Answer 1

Answer -

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• The total voltage across the three resistors is given by, For the circuit shown in Fig 6: ... The direction of the magnetic field produced by a current carrying wire can be determined by.

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Answer 2

Given that,

Current = I

According to figure,

Radius of first arc = r

Radius of second arc = r

Radius of third arc = r

The magnetic field are zero at center O due to both lines A and B because these lines are along magnetic field line.

We need to calculate the magnetic field at center due to first arc

Using formula of magnetic field

$B_{1}=\dfrac{\mu_{0}I\theta}{4\pi r}$...(I)

Put the value into the formula

$B_{1}=\dfrac{\mu_{0}I\theta}{4\pi (3r)}(-k)$

We need to calculate the magnetic field at center due to second arc

Put the value in the equation (I)

$B_{2}=\dfrac{\mu_{0}I\theta}{4\pi (2r)}(k)$

We need to calculate the magnetic field at center due to third arc

Put the value in the equation (I)

$B_{1}=\dfrac{\mu_{0}I\theta}{4\pi (r)}(-k)$

We need to calculate the net magnetic field

Using formula of magnetic field

$B = B_{1}+B_{2}+B_{3}$

Put the value into the formula

$B=-\dfrac{\mu_{0}I\theta}{12\pi r}+\dfrac{\mu_{0}I\theta}{8\pi r}-\dfrac{\mu_{0}I\theta}{4\pi r}$

$B=\dfrac{5\mu_{0}I\theta}{24\pi r}$

Hence, The magnetic field at center is $\dfrac{5\mu_{0}I\theta}{24\pi r}$

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Topic : magnetic field

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