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Chapter 27: Problem 15

A conducting loop of area \(240 \mathrm{cm}^{2}\) and resistance \(12 \Omega\) is perpendicular to a spatially uniform magnetic field and carries a 320 -mA induced current. At what rate is the magnetic field changing?

Short Answer

Expert verified
The magnetic field is decreasing at a rate of 160 T/s.

Step by step solution

01

Convert units

The area of the loop is given in cm², but in SI units, it should be in m². It can be converted to m² using the conversion factor 1 m² = 10,000 cm². Hence, Area \(A = 240 / 10000 = 0.024 m² \)
02

Apply Faraday's Law

According to Faraday's law of electromagnetic induction, the induced emf in a coil is equal to the rate of change of magnetic flux through it. The emf can be rated as \( \epsilon = - d \Phi/dt \)
03

Relate emf to resistance and current

Ohm's law states that the emf can also be considered as the product of the resistance of the conductor and the induced current passing through it. Therefore, \( \epsilon = IR = 12Ω * 0.32A = 3.84 V \)
04

Calculate the rate of flux change

Faraday's law states that the emf is equal to the rate of change of magnetic flux. Therefore, \( - d \Phi/dt = \epsilon \) so \( d \Phi/dt = - \epsilon = -3.84 Vs^{-1} \).
05

Calculate the rate of change of the magnetic field

The magnetic flux Φ is given by \( \Phi = BA \), where B is the magnetic field and A is the area of the loop. Hence, \( dB/dt = (1/A) * d \Phi/dt = (-3.84/0.024) T/s = -160 T/s \). Since the magnetic field is decreasing (as indicated by the negative sign), the rate of change of the magnetic field is 160 Teslas per second.

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Most popular questions from this chapter

A single-turn loop of radius \(R\) carries current \(I\). How does the magnetic- energy density at the loop center compare with that of a long solenoid of the same radius, carrying the same current, and consisting of \(n\) turns per unit length?

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