A 243-turn circular coil of radius 30 cm is placed in a spatially uniform magnetic field with its axis parallel to the magnetic field lines. The magnetic field is changing temporally at a rate of -0.04 T/s i.e., dB/dt = -0.04 T/s.

The resistance of the coil is 6.3 Ω.

What is the magnitude of the current circulating through the coil?

Neglect the effect of Earth’s magnetic field.

Calculate your answer in the units of ampere (A).

Write your answer in decimal form with two digits to the right of the decimal point (e.g. 5.32); do not write any units.

Question

A 243-turn circular coil of radius 30 cm is placed in a spatially uniform magnetic field with its axis parallel to the magnetic field lines. The magnetic field is changing temporally at a rate of -0.04 T/s i.e., dB/dt = -0.04 T/s.

The resistance of the coil is 6.3 Ω.

What is the magnitude of the current circulating through the coil?

Neglect the effect of Earth’s magnetic field.

Calculate your answer in the units of ampere (A).

Write your answer in decimal form with two digits to the right of the decimal point (e.g. 5.32); do not write any units.

BlackTom AI · Accepted Answer

Let’s begin by identifying the key quantities involved in this induction problem. The coil has N = 243 turns and radius r = 0.30 m, so its cross-sectional area is A = π r^2 = π × (0.30)^2 = π × 0.09 ≈ 0.282743 m^2.

Because the coil’s axis is parallel to the magnetic field lines, the magnetic flux through the coil is Φ = B A (the normal to the coil is aligned with B), so the rate of change of flux is dΦ/dt = A dB/dt. If the magnetic field changes at a rate of dB/dt = -0.04 T/s, the magnitude of the induced electromotive force (emf) in a coil with N turns is |ε| = N |dΦ/dt| = N A |dB/dt|.

Plugging in the numbers: |ε| = 243 × 0.282743 × 0.04. To simplify the multiplication, multiply A by dB/dt first: 0.282743 × 0.04 ≈ 0.0113097. Then multiply by N: 0.0113097 × 243 ≈ 2.748 V (units of volts, but we are focusing on magnitude for current calculation).

The coil has a resistance R = 6.3 Ω. By Ohm’s law, the induced current magnitude is I = |ε| / R = 2.748 / 6.3. Performing this division gives I ≈ 0.436 A.

In summary, the sequence is: compute area from radius, relate emf to dB/dt via N and area, then apply Ohm’s law with the given resistance to obtain the current. This approach uses Faraday’s law for a tightly wound, single-loop-like coil and the basic resistance relation, ensuring units are consistent throughout the steps.

Note: Throughout the calculation, only the magnitude of the emf and current is considered, since the question asks for the magnitude of the current.

ENGR-217:201,202,203,204,501,502,505,506,515,516,521,522,523,524,525,526,535,536,PHYS-217:203,204,501,502,503,531,532,533 Final Exam

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