Question

A solenoid of length 0.6 m has a radius of 2 cm and is made up of 600 turns. If it carries a current of 4 A, then the magnitude of the magnetic field inside the solenoid is

• A. 6.024 × 10-3 T
• B. 8.24 × 10-3 T
• C. 5.024 × 10-3 T
• D. 7.024 × 10-3 T

Answer 1

Answer: (C)

5.024 × 10-3 T

Answer 2

Here $\mathrm { n } = \frac { 600 } { 0.6 } = 1000$ turns $/ \mathrm { m } , \mathrm { I } = \mu \mathrm { A }$

$1 = 0.6 \mathrm {~m} , \mathrm { r } = 0.02 \mathrm {~m}$

$\because \frac { 1 } { \mathrm { r } } = 30$ i.e. $1 > > \mathrm { r }$

Hence we can use long solenoid formula then

$\therefore B = \mu _ { 0 } n \mathrm { I } = 4 \pi \times 10 ^ { - 7 } \times 10 ^ { 3 } \times 4 = 50.24 \times 10 ^ { - 4 }$

$= 5.024 \times 10 ^ { - 3 } \mathrm {~T}$