A straight wire is carrying a current of 2.0 A.It is placed at an angle of 60° with respect to a 0.20-T $\vec { B }$ field.If magnetic force exerted on the wire has a magnitude of 0.40 N,what is the length of the wire?

Question

Quizplus · Accepted Answer

We know that magnetic force experienced by the wire is given by:

$F_B = ILB\sin	heta$

where I is the current, L is the length of the wire, B is the magnetic field, and $	heta$ is the angle between the wire and the magnetic field.

In this case, we have:

$I = 2.0 A$
$B = 0.20 T$
$	heta = 60°$
$F_B = 0.40 N$

Substituting these values, we get:

$0.40 N = (2.0 A)(L)(0.20 T)\sin(60°)$

Solving for L, we get:

$L = \frac{0.40 N}{(2.0 A)(0.20 T)\sin(60°)} = 1.2 m$

Therefore, the length of the wire is 1.2 m, which is answer choice B.

A Straight Wire Is Carrying a Current of 2 B⃗\vec { B }B

A Straight Wire Is Carrying a Current of 2 $\vec { B }$