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If an electric charge +q moves with a velocity v through a magnetic field B then magnetic field apply force F_{m}on the charge. Consider a magnetic field acting along the y-axis, +q charge moves along XY making angleθ with the magnetic field. In such case, charge experiences a magnetic force acting along z-axis i.e. perpendicular to the plane of v and B as shown in Fig. Experimentally it is found that magnitude of magnetic force is,
(i) directly proportional to the magnitude of the charge
$$i.e. \;F_m∝ q..(i)$$
(ii) directly proportional to the strength of the magnetic field
$$i.e.\;F_m∝ v..(ii)$$
(iii) directly proportional to the sine of the angle between v and B
$$i.e.\;F_m∝ B..(iii)$$
(iv) directly proportional to the sine of the angle between v and B
$$i.e.\;F_m∝sinθ$$
Combining equation (i), (ii), (iii) and (iv) we get,
$$F_m∝ qvBsinθ$$
$$or,\;F_m=kqvBsinθ$$
Where k is proportionality constant and it values is 1 so we obtain
$$Therefore,\;F_m=qvBsinθ$$
In vector form,
$$\overrightarrow{F_m}=q(\overrightarrow{v}x\overrightarrow{B})$$
This force is also called Lorentz magnetic force.
By this it is proved that F_{m }is perpendicular to the plane containing v and B.
Direction of Magnetic Force
We already know that magnetic force F_{m }is perpendicular to the plane containing v and B. The force could be up or down as shown in Fig. ,since both directions are perpendicular to the plane containing v and B. Fleming left-hand rule illustrated below is used to determine a direction of the magnetic force.
Fleming’s left-hand rule
According to this rule, the forefinger, middle finger and thumb of the left hand are stretched mutually perpendicular to each other in such way that if the forefinger points the direction of field (B) and the middle finger points in the direction of motion of charge particle then the direction of thumb gives the direction of the force as shown in Fig. This rule can be also used if v is not perpendicular to B.
Special cases: Consider an electric charge +q moving with velocity \(\overrightarrow{v} \) through a magnetic field \(\overrightarrow{B}\). Then the magnetic force \(\overrightarrow{F_m}\) on the charge is given by:
Magnitute of force, \(F_m=qvBsinθ\)
(i) When θ = 0^{o}or 180^{o,}sinθ = 0, so F_{m}= qvB(0) = 0
Hence the charged particle moving parallel or antiparallel to the direction of magnetic field experiences no force.
(ii) Whenθ = 90^{o ,}sinθ = 1, so F_{m}= qvB
Hence force experienced by a charged particle is maximum when it is moving perpendicular to the direction of magnetic field.
(iii) When v = 0, i.e. charged particle is at rest.
F_{m}= q (0) B sinθ = 0 If a charged particle is at rest in a magnetic field , it experiences no force.
(iv) When q = 0, F_{m}= 0
It means electrically neutral particle moving in a magnetic field experiences no force.
Unit of B:
The SI-unit of magnetic field strength is 1 tesla (i.e.1 T)
We get,
$$F_m=qvBsinθ$$
$$or\;B=\frac{F_m}{qvsinθ}$$
when q = 1C, v = 1ms^{-1},θ = 90^{o} and F_{m }= 1N, then
$$B=\frac{1}{1\times1\times1}=1T$$
Hence magnetic field strength aat a point 1T, if 1C charge moving with 1ms^{-1}at right angles to the magnetic field, experiences 1N force at that point.
Note that magnetic field of earth at its surface is about 10^{-4}T.
Dimension of B:
We have,
$$B= \frac{F_m}{qvsinθ}$$
Dimension of B
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jwakin james
An electric charge q is moving with a velocity v in the direction of magnetic field B.The magnetic force acting on the charge is
Mar 21, 2017
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