Question

The component of the external magnetic field along the central axis of a 18 18 turn circular coil of radius 27.0 27.0 cm decreases from 1.90 1.90 T to 0.500 0.500 T in 2.90 2.90 s. If the resistance of the coil is R = 4.50 R=4.50 Ω, what is the magnitude of the induced current in the coil?

Answer 1

The magnitude of the induced current in the coil will be "0.44 A". To understand the calculation, check below.

Resistance and Current

According to the question,

Number of turns, N = 18

Radius, r = 27.0 cm

Coil's resistance, R = 4.50 Ω

Time, t = 2.90 s

Initial magnetic field, B₁ = 1.90 T

Final magnetic field, B₂ = 0.500 T

We know that,

The area of cross-section = πr²

By substituting the values,

                                            = 3.14 × (0.27)²

                                            = 0.228 m²

The induced EMF be:

E = $\frac{N\times A\times (B_1-B_2)}{\Delta t}$

By substituting the values,

  = $\frac{18\times 0.228\times (1.9-0.5)}{2.90}$

  = $\frac{18\times 0.228\times 1.4}{2.90}$

  = 1.98 V

hence, The current will be:

→ I = $\frac{E}{R}$

    = $\frac{1.98}{4.5}$

    = 0.44 A

Thus the above response is correct.

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