Question

33. A long solenoid having 200 turns/cm and carries
current i. Magnetic field at its axis is
6.28 x10-2 wb/m². An another solenoid having
having 100 turns/cm and carries - current, then
magnetic field at its axis will be :-
(1) 1.05 x 10-4 Wb/m2
(2) 1.05 x 10-2 Wb/m2
(3) 1.05 x 10-5 Wb/m²
(4) 1.05 x 10-3 Wb/m2

Answer 1

The magnetic field at the second solenoid’s axis will be option (2): 1.05 * 10⁻² Wb/m².

Explanation:

Required Formula:

Magnetic field at the axis of a solenoid is given by, B = µo * n *  i  

For the first long solenoid:

No. of turns of the solenoid, n1 = 200 turns/cm

Current through the solenoid, i1 = i

The magnetic field at the axis of the first solenoid, B1 = 6.28 x10⁻² Wb/m²

Based on the above formula the magnetic field for the first solenoid at its axis will be,

B1 = µo * n1 *  i1  

⇒ 6.28 x10⁻² = µo * 200 *  i …… (i)

For the second solenoid:

No. of turns of the solenoid, n2 = 100 turns/cm

Current through the solenoid, i2 = i/3

Let the magnetic field at the axis of the second solenoid be “B2”.

Based on the above formula the magnetic field for the second solenoid at its axis will be,

B2 = µo * n2 *  i2  

⇒ B2 = µo * 100 *  i/3 …… (ii)

Now, dividing the eq. (ii) by (i), we get

[B2] / [6.28 x10⁻²] = [µo * 100 *  i/3] / [µo * 200 *  i]

⇒ B2 = [6.28 * 10⁻²] / 6

⇒ B2 = 1.05 8 10⁻² Wb/m²

Thus, the magnetic filed of the second solenoid at its axis is 1.05 * 10⁻² Wb/m².

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Answer 2

Explanation:

Given A long solenoid having 200 turns/cm and carries  current i. Magnetic field at its axis is  6.28 x 10^-2 wb/m². An another solenoid having 100 turns/cm and carries - current i/3, then  magnetic field at its axis will be :-

• We know that B1 = μo n1 I1
•   Also           B2 = μo n2I2
• Therefore B2 / B1 = n2 / n1 x I2/I1
•                              = 100 / 200 x i/3 / i
•                                = 1/6
• Now B2 = B1 / 6
•            = 6.28 x 10^-2 / 6
•             = 1.05 x 10^-2 Wb / m^2
• Reference link will be
• https://brainly.in/question/10939348