Question

A tightly wound 90 turn coil of radius 15 cm has a magnetic field 4 × 10-4 Tat its centre. The current flowing through it is

• A. 1.06 A
• B. 2.44 A
• C. 3.44 A
• D. 4.44 A

Answer 1

Answer: (A)

1.06 A

Answer 2

Here, N = 90

$\therefore \mathrm { I } = \frac { 2 \mathrm { RB } } { \mathrm { N } \mu _ { 0 } } = \frac { 2 \times 15 \times 10 ^ { - 2 } \times 4 \times 10 ^ { - 4 } } { 4 \pi \left[ \times 10 ^ { 7 } \times 90 \right. } = 1.06 \mathrm {~A}$