Question

A solenoid of length 100 cm is wound uniformly with 10000 turns of wire. It carries a current of 4A. What is the value of the magnetic field on the axis at the end?

a) 2.0 x 10^(⁻⁴) T
b) 4.0 x 10^(⁻⁴) T
c) 6.0 x 10^(⁻⁴) T
d) 8.0 x 10^(⁻⁴) T

Answer 1

To find the magnetic field on the axis at the end of a solenoid, use the formula B = μ₀nI, where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. Plugging in the given values, the magnetic field is 1.6 × 10⁻² T.

Step-by-step explanation:

To find the magnetic field on the axis at the end of a solenoid, you can use the formula:

B = μ₀nI

where B is the magnetic field, μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), n is the number of turns per unit length, and I is the current. In this case, the solenoid has a length of 100 cm and 10000 turns, so the number of turns per unit length is 100 turns/cm or 1000 turns/m. The current is given as 4A. Plugging these values into the formula, we get:

B = (4π × 10⁻⁷ T·m/A) × (1000 turns/m) × (4A) = 1.6 × 10⁻² T

The value of the magnetic field on the axis at the end of the solenoid is 1.6 × 10⁻² T.