The magnitude of force per unit length on a wire carrying current $I$ at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to $L$ and current in them is $I$ ) is:

Question

The magnitude of force per unit length on a wire carrying current $I$ at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to $L$ and current in them is $I$ ) is:

Toppr Admin · Accepted Answer

Net Magnetic field-xA0-at-xA0-point-xA0-O-xA0-due-xA0-to-xA0-both-xA0-the-xA0-wires-xA0-shall-xA0-beB-x2212-2-x222B-00900-x3BC-0I2-x3C0-lsin-x3B8-d-x3B8-xA0-xA0- along -x2212-k

The magnitude of force per unit length on a wire carrying current I at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to L and current in them is I) is:

Net Magnetic field at point O due to both the wires shall beB=−2∫00900μ0I2πlsinθdθ along −k

The magnitude of force per unit length on a wire carrying current $I$ at O due to two semi infinite wires as shown in figure, (if the distance between the long parallel segments of the wire being equal to $L$ and current in them is $I$ ) is:

Net Magnetic field at point O due to both the wires shall be
$B = - 2 \int_{90^{0}}^{0^{0}} \frac{μ_{0} I}{2 π l} sin θ d θ$ along $- k$