Question

Calculate the magnitude of the magnetic force exerted on a wire that is 20 mmmm long and carries a current of 4.0 AA when it is suspended inside a solenoid at an angle of 45∘∘ to the magnetic field. The solenoid has 800 turns per meter of length and carries a current of 2.0 AA .

Answer 1

Answer:

$1.14\times 10^{-4}N$

Explanation:

We are given that

Length of wire,L=20 mm=$20\times 10^{-3} m$

$1mm=10^{-3} m$

Current,I=4 A

$\theta=45^{\circ}$

Number of turns per unit length,n=800

Current,i=2 A

Magnetic field of solenoid,B=$\mu_0ni=4\pi\times 10^{-7}\times 800\times 2=2.01\times 10^{-3} T$

Where $\mu_0=4\pi\times 10^{-7}$

Magnetic force,F=$IBLsin\theta$

Using the formula

$F=4\times 2.01\times 10^{-3}\times 20\times 10^{-3}sin45$

$F=1.14\times 10^{-4}N$