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The magnetic field around the head has been measured to be approximately3.0×10-8G. Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modelling them as a single circular current loop 16 cm (the width of a typical head) in diameter. What is the current needed to produce such a field at the center of the loop?

Short Answer

Expert verified

The current needed to produce3.0×10-8G at the center of the loop is3.8×10-7A.

Step by step solution

01

The magnetic field at the center of the loop

The magnetic field at the center of the loop is given by:

B=μ0I2R

Where B is the magnetic field due to wire,μ0 is the permeability of the vaccum, l is the current through the wire, and R is the distance from the wire.

02

The calculation of the current flowing in the loop 

Given data:

  • The magnetic field at the center of the loop is 3.0 ×10-12 T.

Using formula :

B=μ0I2RI=2BRμ0

Now, putting the values of constants in the above equation as:

I=2×8×102×3.0×10124×π×107I=3.8×107A

Thus, the current needed to produce3.0×108G at the center of the loop is3.8×107A.

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Most popular questions from this chapter

(See Discussion Question Q25.14.) Will a light bulb glow more brightly when it is connected to a battery as shown in Fig. Q25.16a, in which an ideal ammeter is placed in the circuit, or when it is connected as shown in Fig. 25.16b, in which an ideal voltmeter V is placed in the circuit? Explain your reasoning.

Batteries are always labeled with their emf; for instances an AA flashlight battery is labelled “ 1.5 V ”. Would it also be appropriate to put a label on batteries starting how much current they provide? Why or why not?

An electron moves at 1.40×106m/sthrough a regionin which there is a magnetic field of unspecified direction and magnitude 7.40×10-2T. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle
between the electron velocity and the magnetic field?

An electron at point in figure has a speed v0=1.41×106m/s. Find (a) the magnetic field that will cause the electron to follow the semicircular path from to and (b) The time required for the electron to move fromAtoB.

Copper has 8.5×1022free electrons per cubic meter. A 71.0-cm

length of 12-gauge copper wire that is 2.05 mm in diameter carries 4.85 A of

current. (a) How much time does it take for an electron to travel the length

of the wire? (b) Repeat part (a) for 6-gauge copper wire (diameter 4.12 mm)

of the same length that carries the same current. (c) Generally speaking,

how does changing the diameter of a wire that carries a given amount of

current affect the drift velocity of the electrons in the wire?

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