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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

Consider the circuit of Fig. E25.30. (a)What is the total rate at which electrical energy is dissipated in the 5.0-Ω and 9.0-Ω resistors? (b) What is the power output of the 16.0-V battery? (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery? (d) Show that the power output of the 16.0-V battery equals the overall rate of consumption of electrical energy in the rest of the circuit.

Fig. E25.30.

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