If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Solution and Explanation
Magnetic field strength, B = 0.25 T
Magnetic moment, M = 0.6 T-1
The angle \(\theta\), between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as:
\(\tau\) =\(MB \sin\) \(\theta\)
= 0.6 × 0.25 sin 30\(\degree\)
= 7.5 × 10-2 J
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Concepts Used:
Magnetism & Matter
Magnets are used in many devices like electric bells, telephones, radio, loudspeakers, motors, fans, screwdrivers, lifting heavy iron loads, super-fast trains, especially in foreign countries, refrigerators, etc.
Magnetite is the world’s first magnet. This is also called a natural magnet. Though magnets occur naturally, we can also impart magnetic properties to a substance. It would be an artificial magnet in that case.
Read More: Magnetism and Matter
Some of the properties of the magnetic field lines are:
- The lines and continuous and outside the magnet, the field lines originate from the North pole and terminate at the South pole
- They form closed loops traversing inside the magnet.
- But here the lines seem to originate from the South pole and terminate at the North pole to form closed loops.
- More number of close lines indicate a stronger magnetic field
- The lines do not intersect each other
- The tangent drawn at the field line gives the direction of the field at that point.