Question

7.
The magnitude of magnetic field induction at the
centre o in the figure shown is (Given: R1 =
10 cm, R2 = 20 cm and i = 1 A)
​

Answer 1

Answer:

Solution =

the given R1 = 10cm

R2 = 20 cm

I =1A

so multiply all the simplified no.

so it can be converted into many more solutions

but I am doing it in simple way

x = 10 cm , 20 cm , 1A

10 cm + 20 cm

= 30cm + 1A

so the diameter = 30 cm.

Radius = 2 paye r

30cm = 2× 22 × 1 a

7

= 44 ÷ 7 = 6. 28

solution =6.28

plzz mark me as a brainliest..

Answer 2

Given:

R1 = 10 cm, R2 = 20 cm and i = 1 Ampere.

To find:

Magnetic field intensity at the centre O of the figure.

Calculation:

Magnetic Field Intensity will be contributed only by the semicircular portion of the figure because

• The axis straight wire pass through the centre and hence will contribute zero field Intensity.

$B = \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{\pi i}{(r1)} \bigg \} - \dfrac{ \mu_{0}}{4\pi} \bigg \{ \dfrac{\pi i}{(r2)} \bigg \}$

$= > B = \dfrac{ \mu_{0}\pi i}{4\pi} \bigg \{ \dfrac{1}{r1} - \dfrac{1}{r2} \bigg \}$

$= > B = ( {10}^{ - 7} \times \pi \times 1) \bigg \{ \dfrac{1}{r1} - \dfrac{1}{r2} \bigg \}$

$= > B = ( {10}^{ - 7} \times \pi) \bigg \{ \dfrac{1}{r1} - \dfrac{1}{r2} \bigg \}$

$= > B = ( {10}^{ - 7} \times \pi) \bigg \{ \dfrac{1}{0.1} - \dfrac{1}{0.2} \bigg \}$

$= > B = ( {10}^{ - 7} \times \pi) \bigg \{ 10- 5 \bigg \}$

$= > B = ( {10}^{ - 7} \times \pi) \bigg \{ 5 \bigg \}$

$= > B = 5\pi \times {10}^{ - 7} \: tesla$

So, final answer is:

$\boxed{ \red{ \bold{B = 5\pi \times {10}^{ - 7} \: tesla}}}$