# Find the magnetic field at axis of a long hollow half cylinder

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The magnetic field at the axis of a long half cylinder if current is 6A and 2m be the radius.

I know it's bit easy but still I am having a problem.
I thought the hollow long cylinder as a cluster of thin wires along its surface assuming
a d$\theta$ element and $\theta$ being angle with horizontal I realise that on cos component will come in my equation as the sin component is being cancelled.

dB = $(\mu_0/4\pi)\frac{2.6.d\theta}{2}$

Now what ? Should I integrate it with limits 0 to $\pi$

What is the direction of the current? A diagram would be useful.

1 vote

Let the current be $i$ amperes and radius be $R$.

Take an elementary wire which subtends an angle $d\theta$ at the centre and $\theta$ angle away from the diameter.

Current in that element is $$di=\frac{i.d\theta}{\pi}$$.

Now the magnetic field at the centre will be $$\int \dfrac{\mu_o di \sin(\theta)}{2 \pi R}$$ (along the direction of diameter).

And the magnetic field at the centre will be $$\int \dfrac{\mu_o di \cos(\theta)}{2 \pi R}$$ (along the direction perpendicular to the diameter).

Take the limits of $\theta$ from $0$ to $\pi$. The second integral will evaluate to zero. The value of first integral is the final answer.

P.S: First of all draw a neat diagram and follow the above steps to reach the answer. Use Biot-Savart's law to find the direction of current due to each wire element.

answered Nov 5, 2016 by (160 points)
edited Nov 9, 2016 by Einstein
ok what i mistook in a hurry was that I took the amgle as - θ and so my sin component was being cancelled