6. Find the magnetic field strength at the centre of a solenoid with 5000 turns and current of 5 A. The permeability of free space i
471 x 10-7 H/m.

0.03 T

Explanation:

The magnetic field B at the center of a solenoid is given by B = μ₀Ni/L where

μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, N = number of turns of solenoid = 5000 turns, i = current in solenoid = 5 A and L= length of solenoid. Since we are not given length of solenoid, let us assume it is 1 meter. So, L = 1 m

So, B = μ₀Ni/L

=  4π × 10⁻⁷ H/m × 5000 turns × 5 A/1m

= 4π × 10⁻⁷ H/m × 25000 A-turns/m

= 314159.27 × 10⁻⁷ T

= 0.031415927 T

≅ 0.03 T

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s