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A 20-cm long solenoid consists of 100 turns of a coil of radius r = 3.0 cm. A current of I_0 in the coiled wire produces a magnetic field of B_0 ​​in the middle of the solenoid.
If we wish to produce a field that is 4-times larger, which of the following changes could work?

a) quadruple the number of turns in the 20-cm long solenoid.
b) half the length of the 100-turn solenoid but maintain 100 turns across that length.
c) quadruple the length of the 100-turn solenoid, but also double the radius of the coil.
d) half the length of the solenoid, and double the number of turns to 200 total turns.
e) double the current in the wire, and double the number of turns in the 20-cm long solenoid.
f) cut the radius of the coil to one-half.

Respuesta :

Answer:

Explanation:

length of the solenoid, l = 20 cm = 0.2 m

radius, r = 3 cm

current in the coil = Io

Magnetic field = Bo

The magnetic field due to the current carrying solenoid is

[tex]B_{0}=\mu_{0}nI_{0}[/tex] ..... (1)

where, n is the number of turns per unit length.

so, increasing the number of turns by four times in 20 cm length solenoid, the magnetic field is 4 times. option (a)

And, by taking half the length of the solenoid, and the double of number of turns, the magnetic field due to the solenoid becomes 4 times. option (d)

onyebuchinnaji

If we wish to produce a field that is 4-times larger, double the current in the wire, and double the number of turns in the 20-cm long solenoid.

The given parameters:

  • length of the solenoid, l = 20 cm
  • number of turns of the solenoid, N = 100
  • current in the coil, = Io
  • magnetic field, = Bo

The magnitude of the magnetic field at the center of the solenoid is calculated as follows;

[tex]B = \mu_o nI[/tex]

where;

  • n is the number of turns per length (N/L)

[tex]B = \frac{\mu_o NI}{L} \\\\B_1L_1 = B_2L_2\\\\L_2 = \frac{B_1L_1}{B_2} \\\\L_2 = \frac{B_1 \times L_1}{4B_1} \\\\L _2 = \frac{1}{4} L_1[/tex]

The magnetic field, current and number of turns are directly proportional;

[tex]B = \mu_0 nI\\\\\frac{B_1}{N_1I_1} = \frac{B_2}{N_2I_2} \\\\N_2 I_2 = \frac{B_2}{B_1} N_1I_1\\\\N_2 I_2 = \frac{4B_1}{B_1} N_1I_1\\\\N_2I_2 = 4N_1I_1\\\\N_2I_2 = 2N_1 2I_1[/tex]

Thus, we can conclude that if we wish to produce a field that is 4-times larger, double the current in the wire, and double the number of turns in the 20-cm long solenoid.

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