Derive an expression for the maximum force experienced by a straight conductor of length I, carrying current I and kept in a uniform magnetic field, B.

Solution

Consider a straight conductor PQ of length l, area of cross section A, carrying current I placed in a uniform magnetic field

\vec{B}

.

Suppose the conductor is placed along x-axis and magnetic field acts along y-axis.
Hence, current I flows from end P to Q and electrons drift from Q to P.

Let,

\vec{v_{d}} = drift velocity of electron

- e = charge on each electron

Therefore,

Magnetic Lorentz force on an electron is given by

\overset{⇀}{f} = - e (\vec{v_{d}} \times \vec{B})

[∵ F = q (\vec{v} \times \vec{B})]

If, n is the number of free electrons per unit volume of the conductor, then
total number of free electrons in the conductor will be N = n (AI) = nAl.

$∴$
Total force on the conductor is

$\vec{F} = N \vec{f} = n A l [- e (\vec{v_{d}} \times \vec{B})] = - n A l e (\vec{v_{d}} \times \vec{B}) . . . (i)$

But, the current through a conductor is related with drift velocity by the relation
$I = n A e {\vec{v}}_{d}$

$∴$ Il = n A e v_dl
We represent I $\vec{l}$ as current element vector and, it acts in the direction of flow of current i.e., along OX.
Then we have I $\vec{l}$ and $\vec{v_{d}}$ in opposite directions. So,

$I \vec{l} = - n Ale \vec{v_{d}}$ ...(ii)

Therefore, from (i) and (ii), we have

$\vec{F} = I (\vec{l} \times \vec{B})$

Magnitude of
F = Il B sin θ
When $π = 90 °, then, F_{\max} = B I l$

The direction of force on a current carrying conductor placed perpendicular to the magnetic field is given by Fleming’s left hand rule. If we stretch the fore finger, central finger and the thumb of our left hand mutually perpendicular to each other such that the fore finger points in the direction of magnetic field, central finger in the direction of current, then the thumb gives the direction of force experienced by the conductor.

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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Moving Charges And Magnetism

Derive an expression for the maximum force experienced by a straight conductor of length I, carrying current I and kept in a uniform magnetic field, B.

Some More Questions From Moving Charges And Magnetism Chapter

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