Question

A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5A. A coil of radius 1 cm having 100 turns and a total resistance of 20 Ω is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer and how?

Answer 1

Answer on Question #56229, Physics / Electromagnetism

A long solenoid of radius $2\mathrm{cm}$ has 100 turns/cm and carries a current of 5A. A coil of radius 1 cm having 100 turns and a total resistance of $20\Omega$ is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer and how?

Solution.

$I = 5\mathrm{A}$ –is current through solenoid; $r_s = 2\mathrm{cm} = 0.02\mathrm{m}$ – radius of solenoid;

$r_c = 1\mathrm{cm} = 0.01\mathrm{m}$ – radius of the coil; $N = 100$ – number of coil turns;

$R = 20\Omega$ – is the total resistance of coil; $n = 100\frac{\mathrm{turns}}{\mathrm{cm}} = 10^4\frac{\mathrm{turns}}{\mathrm{m}}$;

$\alpha = 0$ – the angle between axes of coil and solenoid.

$$| q | = \left| \int_ {t 1} ^ {t 2} I d t \right| = \left| \int_ {t 1} ^ {t 2} \frac {\mathrm {E} _ {i n d}}{R} d t \right| = \left| \frac {1}{R} \int_ {t 1} ^ {t 2} \mathrm {E} _ {i n d} d t \right|$$

$E_{ind} = -\frac{d\Phi}{dt}$ – is electromotive force of the induction; $t1, t2$ – moments of time before and after change current direction;

$$| q | = \left| \frac {1}{R} \int_ {t 1} ^ {t 2} \frac {d \Phi}{d t} d t \right| = \frac {1}{R} | \Phi (t 2) - \Phi (t 1) |$$

$\Phi = NBS\cos \alpha$ – the total magnetic flux trough coil; $S = \pi r_c^2$ – area of the coil;

$$| q | = \frac {N S \cos \alpha}{R} | B (t 2) - B (t 1) |$$

Let $B_0$ – magnetic field before current reversed, then $B(t2) = -B_0$, $B(t1) = B_0$

$$| q | = \frac {N S \cos \alpha}{R} | - B _ {0} - B _ {0} | = \frac {2 B _ {0} N S \cos \alpha}{R}$$

$B_{0} = \mu_{0}In$ – magnetic field in solenoid;

$\mu_0 = 4\pi *10^7\mathrm{H / m}$ – magnetic constant;

$$| q | = 2 \mu_ {0} n N S \frac {I}{R} \cos \alpha = 2 \pi \mu_ {0} n N r _ {c} ^ {2} \frac {I}{R} \cos \alpha$$ $$| q | = 2 * 3. 1 4 * 4 * 3. 1 4 * 1 0 ^ {4} * 1 0 ^ {2} * (1 0 ^ {- 2}) ^ {2} * \frac {5}{2 0} * \cos 0 \approx 1. 9 7 * 1 0 ^ {- 4} (\mathrm {C})$$

According to Lenz's law, direction of current in coil is the same as direction of current in solenoid before direction reversed.

Answer: $1.97 * 10^{-4} \, \text{C}$, direction of current in coil is the same as direction of current in solenoid before direction reversed.

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