Question

cm. the current in the solenoid is increasing at a uniform rate of 36.0 a>s. what is the magnitude of the induced electric field at a point near the center of the solenoid and (a) 0.500 cm from the axis of the solenoid; (b) 1.00 cm from the axis of the solenoid?

Answer 1

The induced emf near the solenoid's center has a magnitude of 0 V/m.

At a location 0.500 cm from the solenoid's axis, the induced electric field has an intensity of 8 x 105 V/m.

the settings provided;

• N = 700 turns/m for the solenoid's number of turns,
• and r = 2.5 cm for the wire's radius.
• I = 36 A/s for the solenoid's current.

The following formula is used to determine the size of the induced emf close to the solenoid's center:

B = 4π × 10⁻⁷ × 700 × 36

  = 0.032 T/s

E = r/2(dB/dt)

E = 0/2*(0.032)

E = 0 V/m

The following formula is used to determine the strength of the induced electric field at a position 0.500 cm from the solenoid's axis:

E = r/2(dB/dr)

E = {(0.5 * 10⁻²)/2}*(0.032)

E = 8 × 10⁻⁵ V/m

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