Question

A force exerted on 3m long conductor having a current of 2A in 0.4T magnetic Induction with an angle of 30

Answer 1

Answer:

$\boxed{\mathfrak{Magnetic \ force \ (F_m) = 1.2 \ N}}$

Given:

Current (i) = 2 A

Magnetic induction/field ($\rm |\overrightarrow{B}|$ = 0.4 T

Length of conducting wire ($\rm |\overrightarrow{l}|$) = 3 m

Angle between conducting wire and magnetic field ($\rm \theta$) = 30°

Explanation:

Magnetic force on current carrying conductor:

$\boxed{ \bold{\overrightarrow{F}_m = i(\overrightarrow{l} \times \overrightarrow{B} )}}$

$\rm \implies |\overrightarrow{F}_m | = i | \overrightarrow{l} | | \overrightarrow{B} | sin \theta \\ \\ \rm \implies F_m = ilBsin \theta$

By substituting values in the equation we get:

$\rm \implies F_m = ilBsin \theta \\ \\ \rm \implies F_m = 2 \times 3 \times 0.4 \times sin 30° \\ \\ \rm \implies F_m = \cancel{2} \times 3 \times 0.4 \times \frac{ 1}{ \cancel{2}} \\ \\ \rm \implies F_m = 3 \times 0.4 \\ \\ \rm \implies F_m = 1.2 \: N$

Answer 2

A force exerted on 3m long conductor having a current of 2A in 0.4T magnetic Induction with an angle of 30