Question

16m
Q3. Calculate the value of magnetic field at a distance of 2 cm from a
very long straight wire carrying a current of 5 A. ( 10 = 411 x 10-7Wb/Am)​

Answer 1

Answer:

Explanation:

Given :

• Distance (d) = 2 cm ⟶ 0.02 m
• Current (I) = 0.5 A
• Value of $\bold{(\mu_0)}$ = 4π × 10-⁷ Wb/Am.

To Find :

• The value of magnetic field.

Solution :

Formula of "magnetic field" is given as ;

$\bold{ \green{B = \dfrac{ \mu_0 \times I }{2 \pi \times d} }}$

Now,

$\longrightarrow \tt \: B = \dfrac{4 \pi \times 10 {}^{ - 7} \times 5 }{2 \pi \times 0.02} \\ \\ \\ \longrightarrow \tt \: B = 2 \times 250 \times 10 {}^{ - 7} \\ \\ \\ \longrightarrow \tt \: B = 500 \times 10 {}^{ - 7} \\ \\ \\ \longrightarrow \tt \bold{ \red{ B = 0.00005 \:T }}$

Hence :

The value of magnetic field is 0.00005 T.

Answer 2

$\large\sf{\color{red}{Answer:}}$

0.00005 T

$\large\sf{\color{deepskyblue}{Explanation:}}$

$\large\sf{\color{teal}{Given :}}$

• Distance (d) = 2 cm ⟶ 0.02 m

• Current (I) = 0.5 A

• Value of (μ0) = 4π × 10-⁷ Wb/Am

$\large\sf{\pink{To~Find :}}$

• The value of magnetic field.

$\large\sf{\color{royalblue}{Solution :}}$

Formula of "magnetic field" is given as :

$\bold{\purple{B = \dfrac{ \mu_0 \times I }{2 \pi \times d} }}$

Now,

$\begin{gathered}\longrightarrow \tt \: B = \dfrac{4 \pi \times 10 {}^{ - 7} \times 5 }{2 \pi \times 0.02} \\ \\ \\ \longrightarrow \tt \: B = 2 \times 250 \times 10 {}^{ - 7} \\ \\ \\ \longrightarrow \tt \: B = 500 \times 10 {}^{ - 7} \\ \\ \\ \longrightarrow \tt \bold{ \red{ B = 0.00005 \:T }}\end{gathered}$

The value of magnetic field is 0.00005 T.