A circular loop of radius 10 cm and three long straight wires carry currents of

Question

A circular loop of radius 10 cm and three long straight wires carry currents of I1 = 60 A, I2 = 20 A, I3 = 10 A, and I4 = 10 A, respectively, as shown. Each straight wire is 20 cm from the center of the loop. In Figure 28.1, the y - component of the resultant magnetic field at the center of the loop is closed to: + 15 mu T - 20 mu T + 20 mu T - 17 mu T - 15 mu T A long straight wire on the z - axis carries a current of 6.0 A in the positive direction. A circular loop in the x - y plane, of radius 10 cm, carries a 1.0 A current, as shown. Point P, at the center of the loop, is 25 cm from the z - axis In Figure 28,6d, an electron is projected from P with a velocity of 4.0 times 106 m/s in the negative x - direction. If y - component of the force on the electron is closed to: + 4.0 times 10 - 18 N - 4.0 times 10 - 18 N zero + 2.0 times 10 - 18 N - 2.0 times 10 - 18 N

Explanation / Answer

First, we need to find the strength of the magnetic field @p. We use amperes law to discover the formula for the mag field from a long straight wire is B = I / ( 2r ).

We plug in the values provided:

B = I / ( 2r )

B = ( 4e-7 )( 6 A ) / ( 2( .25 m ))

B = 4.8e-6 T. The direction is UP according to the right hand rule.

Now, to find the mag field @ p from the circle, we use law of biot-savart to find that the formula for mag field at center of circle is B = I / 2R. Lets plug in values:

B = I / 2R

B = ( 4e-7 )( 1 A ) / ( 2( .1 m ))

B = 6.283e-6 T. The direction is OUT of page.

Now, the question is asking for the force on an electron moving left @p. According to another right hand rule, the UP mag field from the wire will create a force OUT OF page. This has no y value to it, but the answer they want is only the y component of the force, so we ignore it.

The OUT mag field from the circle will cause a DOWN y force, so this is the only mag field we will use in our calculation.

We will use F = qvXB, but only use the mag field from the circle. The q of an electron is 1.6e-19 C.

F = ( 1.6e-19 C )(4e6 m/s)( 6.283e-6 T ). The cross product does not pmatter because the mag field is perfectly perpendicular to the velocity.

F = 4.02e-18 N.

Remember, the y component is DOWN, so the correct answer is B) -4e-18 N

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