Login/Sign Up
Home » 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?
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?
CBSE | Class 12 | Moving Charges and Magnetism | NCERT | Physics
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?

The wire’s length (l) is 3 cm or 0.03 m

The current running through it (I) is 10 A.

The magnetic field strength (B) is 0.27 T.

the angle formed by the current and the magnetic field is \theta = {90^ \circ }

the magnetic force exerted on the wire can be calculated as:

F = BIl\sin \theta

Putting the values in the above equation, we get

= 0.27 \times 10 \times 0.03 \times \sin {90^ \circ }

= 8.1 \times {10^{ - 2}}N

As a result, the magnetic force on the wire is 8.1 \times {10^{ - 2}}N. Fleming’s left-hand rule can be used to determine the force’s direction.

i

More Material