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Fundamentals Of Physics
Found in: Page 896

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Short Answer

In Figure (a), a uniform magnetic field increases in magnitude with time t as given by Figure (b), where the vertical axis scale is set by Bs=9.0 m Tand the horizontal scale is set by ts=3.0 s A circular conducting loop of area 8.0×10-4m2lies in the field, in the plane of the page. The amount of charge q passing point A on the loop is given in Figure (c) as a function of t, with the vertical axis scale set by qs=3.0 s and the horizontal axis scale again set by localid="1661854094654" ts=3.0 s. What is the loop’s resistance?

Resistance of the loop is, R = R=0.0012Ω0.0012Ω

See the step by step solution

Step by Step Solution

Step 1: Given

  1. Cross-sectional area A=8×10-4m2
  2. Magnetic field on vertical axis, Bs=9.0 mT=9×10-3 T
  3. Horizontal axis, ts=3.0 s
  4. Vertical axis, qs=6 mC=6×10-3C

Step 2: Determining the concept

Calculate the induced emf by using the slope of the line form Fig. 30-42(b) in Faraday’s law. Find the current by finding the slope of the line from Fig.30-42 (c) to find the resistance of the loop by using Ohm’s law.

Faraday's law of electromagnetic induction states, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Formulae are as follow:


Where, ΦB is magnetic flux, dt is time, dq is charge, i is current, R is resistance, 𝜀 is emf,

Step 3: Determining the resistance of the loop

First, find the emf by using Faraday’s law.

From Fig.30 - 42 (b) the slope of the line i.e. dB/dt can be found as,

dBdt=9×10-3T3.0s=0.003 T/s

Faraday’s law is given as,


As magnetic field is perpendicular to the area vector,



role="math" localid="1661853786477" ε=BAε=dBAdtε=-AdBdt

Substituting the values,

ε=-8.0x10-40.003ε=-0.024×10-4 V

Now, the current can be calculated from Fig. 30- 42 (c).

The slope of the graph gives dq/dt.


i=dqdt=6×10-3C3.0 si=0.002 A

Now, relate the induced emf to resistance and current by using Ohm’s law.


Substituting the values,


Hence, resistance of the loop is, R=0.0012Ω

Therefore, we calculated the resistance of the loop by using the slopes from the graph, Faraday’s law, and Ohm’s law.

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