Question

find net magnetic field at point $p\left(L=10cm\right)$

• A. $5.23\times {10}^{-5}$
• B. $2.23\times {10}^{-5}T$
• C. $2.23T$
• D. None

Answer 1

Answer: (D)

None

Given: l=0.1 m ,i=5A

Solution: This question is solved by using the the concept of current carrying straight conductor

Generally ,the formula used is$B=\frac{\mu _{0}i}{4\pi d}\left ( sin\alpha +sin\beta \right )$

Where d=perpendicular distance from current carrying wire to the point where we have to find the magnetic field

$sin\alpha$= angle subtending from one end at point P

$sin\beta$=angle subtending from other end at point P

Now ,there are 2 vertically wire and 1 horizontally wire

For vertically wire:

$B=\frac{{\mu }_{0}i\times 2}{4\pi l}(sin{0}^0+sin{45}^0)\odot$

Multiplied by 2 because there are 2 wires

$B\left(1\right)+\left(3\right)=2\times \frac{{\mu }_{0}i\times 2}{4\pi l}(sin{0}^0+sin{45}^0)$.....(1)

For horizontally wire:

$B\left(2\right)=\frac{{\mu }_{0}i}{4\pi l}(sin{45}^0+sin{45}^0)\otimes$ ...(2)

Net B=B(1)+B(3)- B(2)

After solving we get :

$B=\frac{\sqrt{2}{\mu }_{0}i}{4\pi l}(2-1)\odot$

B=$7.05\times {10}^{-6}$

Hence the correct option is D