Question

4. Solenoid of length 15 cm has 15000
turns carries a current of 2 A, the
magnitude of magnetic induction along
its axis is
(a) 2pie x 10^-2Wb/m
(c) 6pie x 10^-3 Wb/m
(b) 4pie ×10^-3Wb/m
(d) 8pie x10^-3Wb/m​

Answer 1

Answer:

(d) 8pie x10^-3Wb/m

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Answer 2

Given:

Length of solenoid $\[ = 15cm = 15 \times {10^{ - 2}}m\]$

Number of turns $\[ = 15000\]$

Current $\[ = 2A\]$

To Find: Magnetic induction along the axis of the solenoid

Solution:

Magnetic induction along the axis of a solenoid can be given as

$\[B = \frac{{{\mu _0}nI}}{L}\]$                                ... (i)

Where, $n=$ number of turns

$I=$ current

$L=$ length of solenoid

$\[{\mu _0}\]=$ permeability of free space and its value is $\[4\pi \times {10^{ - 7}}N/{A^2}\]$

Therefore, putting values in equation (i), we have,

$\[B = \frac{{4\pi \times {{10}^{ - 7}}N/{A^2} \times 15000 \times 2A}}{{15 \times {{10}^{ - 2}}m}}\]$

$\[B = 8\pi \times {10^{ - 7}} \times {10^3} \times {10^2}\]$

$\[B = 8\pi \times {10^{ - 2}}Wb/{m^2}\]$

Hence, the magnetic induction along the axis of the solenoid is $\[8\pi \times {10^{ - 2}}Wb/{m^2}\]$.