Question

Two solenoids are nested coaxially such that their magnetic fields point in opposite directions. Treat the solenoids as ideal. The outer one has a radius of 20 mm, and the radius of the inner solenoid is 10 mm. The length, number of turns, and current of the outer solenoid are, respectively, 20.5 cm, 545 turns, and 4.53 A. For the inner solenoid the corresponding quantities are 18.9 cm, 331 turns, and 1.49 A. At what speed, v1,should a proton be travelling, inside the apparatus and perpendicular to the magnetic field, if it is to orbit the axis of the solenoids at a radius of 5.69 mm?V1 =_______m/s And at what speed, v2, for an orbital radius of 16.9 mm?V2 =_________m/s

Answer 1

Answer:

Part a)

$v = 6.4 \times 10^3 m/s$

Part b)

$v = 2.44 \times 10^4 m/s$

Explanation:

Magnetic field due to outer solenoid inside it is given as

$B_1 = \mu_0 \frac{N}{L} i$

$B_1 = (4\pi \times 10^{-7})(\frac{545}{0.205})(4.53)$

$B_1 = 0.015 T$

Now similarly we have

magnetic field due to inner solenoid on it axis is given as

$B_2 = \mu_0 \frac{N}{L} i$

$B_2 = (4\pi \times 10^{-7})(\frac{331}{0.189})(1.49)$

$B_2 = 3.28\times 10^{-3} T$

Part a)

Now if a proton is revolving around the common axis of two solenoids

then we will have

$qv(B_1 - B_2) = \frac{mv^2}{R}$

$(1.6 \times 10^{-19})(15 - 3.28)\times 10^{-3} = \frac{1.66\times 10^{-27} v}{5.69 \times 10^{-3}}$

$v = 6.4 \times 10^3 m/s$

Part b)

Now proton is revolving in the field of outer cylinder only

so again we have

$qvB_1 = \frac{mv^2}{R}$

$(1.6 \times 10^{-19})(15)\times 10^{-3} = \frac{1.66\times 10^{-27} v}{16.9 \times 10^{-3}}$

$v = 2.44 \times 10^4 m/s$

Answer 2

The speed of the proton inside the apparatus is 6,389.2 m/s.

The speed of the proton at the orbital radius is 2.43 x 10⁴ m/s.

Magnetic field due to outer solenoid

The magnetic field due to the outer solenoid is calculated as follows;

$B_i = \frac{\mu_0 Ni}{L} \\\\B_i = \frac{4\pi \times 10^{-7} \times 545 \times 4.53}{0.205} \\\\B_i = 0.015 \ T$

Magnetic field due to inner solenoid

$B_f = \frac{\mu_0 Ni}{L} \\\\B_f = \frac{4\pi \times 10^{-7} \times 331 \times 1.49}{0.189} \\\\B_f = 3.28 \times 10^{-3} \ T$

Velocity of proton inside the apparatus;

$qv\Delta B = \frac{mv^2}{r} \\\\v= \frac{qr\Delta B }{m} \\\\v = \frac{1.6 \times 10^{-19} \times 5.69 \times 10^{-3} (0.015 - 3.28 \times 10^{-3})}{1.67 \times 10^{-27}} \\\\v = 6,389.2 \ m/s$

Velocity of the proton at the outer shell

$v= \frac{qrB_i}{m} \\\\v = \frac{1.6 \times 10^{-19} \times 16.9 \times 10^{-3} \times 0.015}{1.67 \times 10^{-27}} \\\\v = 2.43 \times 10^4 \ m/s$

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