ROSHAN MUJEEB
Last Activity: 4 Years ago
Let us take an element of thickness dx at a distance x from
origin on the wire OC. Magnetic fieldBABAproduced at P(x,0,0)
due to wires placed at A and B are
BA=μ0I/2πR,BB=μ0I/2πRBA=μ0I/2πR,BB=μ0I/2πR
Component ofBAandBBBAandBBalong x-axis cancel, while those along
y-axis add up to give total field,
B=2(μ0I2πR)cosθ=2μ0I2πRxR=μ0Iπx(a2+x2)B=2(μ0I2πR)cosθ=2μ0I2πRxR=μ0Iπx(a2+x2)
The force dF acting on the current element isdF→=I(dl→×B→)dF→=I(dl→×B→)
dF=μ0I2πxdxa2+x2[∵sin90∘=1]dF=μ0I2πxdxa2+x2[∵sin90∘=1]
Hence, net force on wire OC,F=∫dFF=∫dF
i.e.,F=μ0I2π∫L0xdxa2+x2=μ0I2πIna2+L2a2F=μ0I2π∫0Lxdxa2+x2=μ0I2πIna2+L2a2along z-direction.if the current in B is reserved, the magnetic field due to the two
wires would be only x-direction and the force on hte current
carrying wire along x-direction will be zero.