CONCEPT: If the circular loop is CONSIDERED as a magnetic dipole, then the dipole moment is the product of current and area. Therefore, the magnitude of magnetic moment = area × current M = IAWhere M = Magnetic moment of copper coil, I = Current flowing through a loop and A = Area of coilEXPLANATION:If e is the charge on an electron revolving in an orbit of radius r with uniform angular velocity ω, then EQUIVALENT current is\(\Rightarrow I = \frac{charge (e)}{time (T)}\)Where T = the period of revolution of electron\(\Rightarrow T=\frac{2\pi r}{v}\)∴ The equivalent current will be \(\Rightarrow I = \frac{ev}{2\pi r}\)Area of the orbit is\(\Rightarrow A= \pi r^2\)The magnitude of the magnetic moment\(\Rightarrow M = \frac{ev}{2\pi r}\times\pi r^2=\frac{evr}{2}\)Multiply and divide the above equation by m, we GET\(\Rightarrow M=\frac{emvr}{2m} = \frac{evr}{2}\) "> CONCEPT: If the circular loop is CONSIDERED as a magnetic dipole, then the dipole moment is the product of current and area. Therefore, the magnitude of magnetic moment = area × current M = IAWhere M = Magnetic moment of copper coil, I = Current flowing through a loop and A = Area of coilEXPLANATION:If e is the charge on an electron revolving in an orbit of radius r with uniform angular velocity ω, then EQUIVALENT current is\(\Rightarrow I = \frac{charge (e)}{time (T)}\)Where T = the period of revolution of electron\(\Rightarrow T=\frac{2\pi r}{v}\)∴ The equivalent current will be \(\Rightarrow I = \frac{ev}{2\pi r}\)Area of the orbit is\(\Rightarrow A= \pi r^2\)The magnitude of the magnetic moment\(\Rightarrow M = \frac{ev}{2\pi r}\times\pi r^2=\frac{evr}{2}\)Multiply and divide the above equation by m, we GET\(\Rightarrow M=\frac{emvr}{2m} = \frac{evr}{2}\) ">

The magnetic dipole moment of a revolving electron is ____________. (e is charge on an electron, v is orbital speed and r is radius of orbit)

Current Affairs General Awareness in Current Affairs 9 months ago

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CONCEPT: If the circular loop is CONSIDERED as a magnetic dipole, then the dipole moment is the product of current and area. Therefore, the magnitude of magnetic moment = area × current M = IAWhere M = Magnetic moment of copper coil, I = Current flowing through a loop and A = Area of coilEXPLANATION:If e is the charge on an electron revolving in an orbit of radius r with uniform angular velocity ω, then EQUIVALENT current is\(\Rightarrow I = \frac{charge (e)}{time (T)}\)Where T = the period of revolution of electron\(\Rightarrow T=\frac{2\pi r}{v}\)∴ The equivalent current will be \(\Rightarrow I = \frac{ev}{2\pi r}\)Area of the orbit is\(\Rightarrow A= \pi r^2\)The magnitude of the magnetic moment\(\Rightarrow M = \frac{ev}{2\pi r}\times\pi r^2=\frac{evr}{2}\)Multiply and divide the above equation by m, we GET\(\Rightarrow M=\frac{emvr}{2m} = \frac{evr}{2}\)

Posted on 06 Nov 2024, this text provides information on Current Affairs related to General Awareness in Current Affairs. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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