The figure shows a part of an electric circuit. The wires AB,CD and EF are long and have identical resistances. The separation between the neighboring wires is 1.0cm . The wires EA and BF have negligible resistance and the ammeter reads 30 A. Calculate the magnetic force per unit length on AB and CD.
A current of $$I=30A$$ passes through ammeters which gets distributed in the three wires As resistance of each wire are equal, thin current gets distributed to the three wires. So, current in each $$AB,CD,EF=\dfrac{30}{3}=10A$$
Magnetic force per length on $$AB$$ will be :-$$[I_{AB}=I_{CD}=I_{EF}=10 A]$$
$$B_1=\dfrac{\mu_{0} I_{AB}I_{CD}}{2\pi r}+\dfrac{\mu_0I_{AB}I_{EF}}{2\pi(2r)}$$
$$=\dfrac{3}{4}\dfrac{\mu oI^2}{\pi r}=\dfrac{3}{4}\times \dfrac{4\pi \times 10^{-7}\times 10^2}{\pi\times 10^{-2}}=3\times 10^{-3}T=3mN/m$$
and that on $$CD$$,
Notice that, the magnetic force on $$CD$$ by $$AB$$ and $$EF$$ are equal but in opposite directions, no force on $$CD=O$$
A beam of electrons is moving with constant velocity in a region having simultaneous perpendicular electric and magnetic fields of strength 20V/m and 0.5T respectively at right angles to the direction of motion of the electrons. Then the velocity of electrons must be
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State Fleming's right hand rule.
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