Question

A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire when it is at an angle of 30° with respect to the field?

Answer 1

Answer:

Magnetic force, F = 0.24 N

Explanation:

It is given that,

Current flowing in the wire, I = 4 A

Length of the wire, L = 20 cm = 0.2 m

Magnetic field, B = 0.6 T

Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

$F=ILB\ sin\theta$

$F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)$

F = 0.24 N

So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.

Answer 2

Explanation:

The given data is as follows.

          length = 20 cm,      current = 4 A

          B = 0.6 T,        $\theta$ = 30°

Hence, formula to calculate the force acting on wire is as follows.

               F = $IlBsin \theta$

Now, putting the given values into the above formula as follows.

             F = $IlBsin \theta$

                = $4 A \times 20 cm \times \frac{10^{-2}}{1 cm} \times 0.6 T \times Sin(30^{o})$

                = 0.24 N

                = 0.2 N

thus, we can conclude that force on the wire is 0.2 N.