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Solenoid

Two coaxial s...

Question

Two coaxial solenoids 1 and 2 of the same length are set so that one is inside the other. The number of turns per unit length are $n_{1}$ and $n_{2}$ . The currents $i_{1}$ and $i_{2}$ are flowing in opposite directions. The magnetic field inside the inner coil is zero. This is possible when

A

i_{1} \neq i_{2}

and

n_{1} = n_{2}

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B

i_{1} = i_{2}

and

n_{1} \neq n_{2}

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C

i_{1} \neq i_{2}

and

n_{1} = n_{2}

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D

i_{1} n_{1} = i_{2} n_{2}

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Solution

The correct option is D $i_{1} n_{1} = i_{2} n_{2}$
For magnetic field inside the solenoid to be zero,
$B_{1} = B_{2}$
Here, $B_{1}$ and $B_{2}$ are the magnetic fields inside the inner cores dues to solenoids 1 and 2 respectively.

$μ_{0} n_{1} i_{1} = μ_{o} n_{2} i_{2}$
$n_{1} i_{1} = n_{2} i_{2}$

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Similar questions

Q. Two coaxial solenoids 1 and 2 of the same length are set so that one is inside the other. The number of turns per unit length are

n_{1}

n_{2}

i_{1}

i_{2}

are flowing in opposite directions. The magnetic field inside the inner coil is zero. This is possible when

Q. Two coaxial long solenoids of equal lengths have current,

i_{1}

i_{2}

, number of turns per unit length

n_{1}, n_{2}

r_{1}, r_{2}

respectively. If

n_{1} i_{1} = n_{2} i_{2}

and the two solenoids carry current in opposite sense, the magnetic energy stored per until length is

[r_{2} > r_{1}]

Q. (a) State Ampere's circuital law, expressing it in the integral form.
(b) Two long coaxial insulated solenoids,

S_{1}

S_{2}

of equal lengths are wound one over the other. A steady current "I" flow through the inner solenoid

S_{1}

to the other end B, which is connected to the outer solenoid

S_{2}

through which the same current "I" flows in the opposite direction so as to come out at end A. If

n_{1}

n_{2}

are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point (i) inside on the axis and (ii) outside the combined system.

Q. Two parallel long wires carry currents

i_{1}

i_{2}

i_{1} > i_{2}

. When the currents are in the same direction, the magnetic field midway between the wires is

10 μ T

. When the direction of

i_{2}

is reversed, it becomes

40 μ T

. Then, the ratio of

i_{1} / i_{2}

Q. Write the following steps in a sequential order to find the current and voltage in the secondary coil. The ratio of the number of turns of a primary to a secondary coil, current through the primary coil and input power are given.
(a) Let the given data be taken as

\frac{N_{1}}{N_{2}}

i_{1}

, (input current) and

P

(b) Substitute the given values of

\frac{N_{1}}{N_{2}}

i_{1}

, in the above formula and get the value of

i_{2}

(c) Write the relation between turn-ratio and the currents as

\frac{N_{1}}{N_{2}} = \frac{i_{2}}{i_{1}}

i_{2}

is the current in the secondary coil.

(d) Write the relation between input power and output power in terms of

V_{1}, i_{1}, V_{2}, i_{2}

V_{1} i_{1} = V_{2} i_{2} = P

V_{1}

V_{2}

are the voltage in the primary and secondary coils respectively.

(e) Substitute the values of

i_{2}

and P in the above relation to get the value of

V_{2}

.

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