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

Step-by-step 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 * 20 cm * (10^(-2))/(1 cm) * 0.6 T * Sin(30^(o))`

= 0.24 N

= 0.2 N

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

Answer 2

Magnetic force, F = 0.24 N

Step-by-step 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* 0.2\ m* 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.