- Introduction
- |
- The Magnetic Field
- |
- Lorentz Force
- |
- Motion of Charged Particle in The Magnetic Field
- |
- Cyclotron
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- Magnetic force on a current carrying wire
- |
- Torque on a current carrying rectangular loop in a magnetic field

- We know that current flowing in a conductor is nothing but the drift of free electron's from lower potential end of the conductor to the higher potential end

- when a current carrying conductor is placed in a magnetic field ,magnetic forces are exerted on the moving charges with in the conductor

- Equation -1 which gives force on a moving charge in a magnetic field can also be used for calculating the magnetic force exerted by magnetic field on a current carrying conductor (or wire)

- Let us consider a straight conducting wire carrying current I is placed in a magnetic field B(x).Consider a small element dl of the qire as shown below in the figure

- Drift velocity of electrons in a conductor and current I flowing in the conductor is given by
I=neAv
_{d}

Where A is the area of cross-section of the wire and n is the number of free electrons per unit volume

- Magnetic force experienced by each electron in presence of magnetic field is

**F**=e(**v**X_{d}**B**)

where e is the amount of charge on an electron

- Total number of electron in length dl of the wire

N=nAdl

- Thus magnetic force on wire of length dl is

**dF**=(nAdl)(e**v**X_{d}**B**)

if we denote length dl along the direction of current by the vector**dl**the above equation becomes

**dF**=(nAev_{d})(**dl**X**B**)

or**dF**=I(**dl**X**B**) -- (12)

where the quantity I**dL**is known as current element

- If a straight wire of length l carrying current I is placed in a uniform magnetic field then force on wire would be equal to

**dF**=I(**L**X**B**) -- (13)

- Direction of force is always perpendicular to the plane containing the current element I
**dL**and magnetic field**B**

- Direction of force when current element I
**dL**and**B**are perpendicular to each other can also be find using either of the following rules**i) Fleming'e left hand rule:-**

If fore finger ,the middle finger and thumb of the left hand are stretched in such a way that the all are mutually perpendicular to each other then,if the fore finger points in the direction of the field (B) and middle finger points in the direction of current I ,the thumb will point in the direction of the force

**ii) Right hand palm Rule:**

Stretch the finger and thumb of the right hand so that they are perpendicular to each other .If the fingers point in the direction of current I and the palm in the direction of field B then the thumb will point in the direction of force

Class 12 Maths Class 12 Physics

- NCERT Solutions: Physics 12th
- NCERT Exemplar Problems: Solutions Physics Class 12
- Concepts of Physics - Vol. 2 HC Verma
- Dinesh New Millennium Physics Class -XII (Set of 2 Vols) (Free Complete Solutions to NCERT Textbook Problems & NCERT Exemplar Problems in Physics-XII)
- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- CBSE Chapterwise Questions-Solutions Physics, Class 12
- CBSE 15 Sample Question Paper: Physics for Class 12th

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