Question

0.8 m long solenoid has 400 turns and a field density of 3.52 x 10 -3T at its centre. Find the current in the solenoid

Answer 1

Answer :

• Magnetic field density = 3.52 × 10⁻³ T
• Length of the solenoid = 0.8 m
• Number of turns = 400
• Current in the solenoid = ?

$\displaystyle\underline{\bigstar\:\textsf{According to the Question :}}$

$\sf:\implies B = \dfrac{\mu_0NI}{l}$

• B = 3.52 × 10⁻³ T

$\sf:\implies 3.52 \times 10^{-3} = \dfrac{4\pi \times 10^{-7}\times 500\times I}{0.5}$

$\sf:\implies 3.52 \times 10^{-3} = \dfrac{4\times 3.14 \times 10^{-7}\times 500\times I}{0.5}$

$\sf:\implies I = \dfrac{3.52 \times 10^{-3}\times 0.5}{4\times 3.14 \times 10^{-7}\times 500}$

$\sf:\implies\underline{\boxed{\pink{\mathfrak{i = 2.59 \ a}}}}$

∴ So the current in the circuit is 2.59 A

Answer 2

Question

0.8 m long solenoid has 400 turns and a field density of 3.52 x 10 -3T at its centre. Find the current in the solenoid

Answer.

B=µ_0 NT

B=3.52×10^-3T

✑3.52×10^-3=4π×500×1/0.5

✑3.52×10^-3

☛4×3.14×10^-7×500×I/0.5

» I =3.52×10^-3×0.5/4×3.14×10^-7×500

◇ I=2.59a ◇