A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is: (μo = 4π × 10-7 T m A-1)

Question

A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is: (μo = 4π × 10-7 T m A-1) (A) 6.28 × 10-4T (B) 3.14 × 10-4T (C) 6.28 × 10-5T (D) 3.14 × 10-5T

Tardigrade Admin · Accepted Answer

Magnetic field at centre of solenoid =μ0nI n=(N/L) =(100/50 × 10-2) =200 tums/m I = 2.5 A On putting the values, B=4π× 10-7× 200 × 2.5 =6.28 × 10-4 T

Q. A long solenoid of $50 c m$ length having $100 t u r n s$ carries a current of $2.5 A$ . The magnetic field at the centre of the solenoid is :
$(μ_{o} = 4 π \times 1 0^{- 7} T m A^{- 1})$

$6.28 \times 1 0^{- 4} T$

$3.14 \times 1 0^{- 4} T$

$6.28 \times 1 0^{- 5} T$

$3.14 \times 1 0^{- 5} T$

Magnetic field at centre of solenoid $= μ_{0} n I$
$n = \frac{N}{L}$
$= \frac{100}{50 \times 1 0 ^{- 2}}$
$= 200$ tums/m
$I = 2.5 A$
On putting the values,
$B = 4 π \times 1 0^{- 7} \times 200 \times 2.5$
$= 6.28 \times 1 0^{- 4} T$

Q. A long solenoid of 50cm length having 100turns carries a current of 2.5A. The magnetic field at the centre of the solenoid is : (μo​=4π×10−7TmA−1)

6.28×10−4T

3.14×10−4T

6.28×10−5T

3.14×10−5T