Question

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T.

What is the magnetic force on the wire?​

Answer 1

Explanation:

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Ans.

Length of the wire, 1= 3 cm = 0.03 m

Current flowing in the wire, I = 10 A

Magnetic field, B = 0.27 T

Angle between the current and magnetic field,

é = 90°

é = 90°Magnetic force exerted on the wire is given as:

F= BIl sin 0.27 x10x0.03 sin90°

0.27 x10x0.03 sin90°8.1x102 N

Hence, the magnetic force on the wire is 8.1-102N.

The direction of the force can be obtained from Fleming's left hand rule.

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Answer 2

Answer:

Magnetic force (F) = 0.081 N

Explanation :

Given :

Length of wire (L) = 3 cm = 3×10^-2 m

Current (I) = 10 A

Magnetic field = B = 0.27 T

Theta = 90° ............(perpendicular)

To find :

Magnetic force (F) = ?

Solution :

We have,

Magnetic force (F) = IL × B = ILBsin(theta)

=> F = 10 × 3×10^-2 × 0.27 × sin90°

=> F = 0.81 × 10^-1× 1

=> F = 0.081 N

Thus, the magnetic force on the wire is 0.081 N