A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Solution

Given,
Current carried by wire, I = 0.40 A,
Radius of circular wire, r = 8.0 cm = $8 \times 10^{- 2} m$
Number of turns in the coil, $n = 100$

Using the formula, we get the magnetic field at the centre of coil as

$B = \frac{μ_{0} nI}{2 r} = \frac{4 π \times 10^{- 7} \times 100 \times 0.4}{2 \times 8.0 \times 10^{- 2}} T$
$= 3.1 \times 10^{- 4} T .$

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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Moving Charges And Magnetism

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Some More Questions From Moving Charges And Magnetism Chapter

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