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alekssr [168]
4 years ago
6

A solenoid with 300 turns has a radius of 0.040 m and is 40 cm long. If this solenoid carries a current of 12 A, what is the mag

nitude of the magnetic field near the center of the solenoid? (μ 0 = 4π × 10-7 T · m/A)
Physics
1 answer:
dalvyx [7]4 years ago
5 0

Answer:

The magnitude of the magnetic field near the center of the solenoid is 11.3 mT

Explanation:

No. of turns N = 300

Radius r = 0.040 m

Length of solenoid L= 40 \times 10^{-2} m

Current I = 12 A

For finding the magnetic field near the center of the solenoid is given by,

   B = \frac{\mu _{o} NI }{L}

Where \mu _{o} = 4\pi  \times 10^{-7}

   B = \frac{4 \pi  \times 10^{-7} \times 300 \times 12}{40 \times 10^{-2} }

   B  = 11.3 \times 10^{-3} T

   B = 11.3 mT

Therefore, the magnitude of the magnetic field near the center of the solenoid is 11.3 mT

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