Q16. An insulating rod of length L carries charge q distributed uniformly on it. The rod is pivoted

Q: What is the correct answer for CUET UG Physics Q16?

The correct answer is b. Consider an elemental length dl at a distance l from the pivot. The charge on this element is dq = (q/L)dl. When this element rotates at a frequency f, the current produced by this rotating charge element is di = dq * f = (q/L)f dl. The magnetic moment (dM) due to this current loop is di * A, where A is the area of the loop formed by the rotation. The area swept by the element at radius l is A = π

Q: Q16. An insulating rod of length L carries charge q distributed uniformly on it. The rod is pivoted at one end and is rotated at a frequency f about a fixed perpendicular axis. Then the magnetic…

The correct answer is b. Consider an elemental length dl at a distance l from the pivot. The charge on this element is dq = (q/L)dl. When this element rotates at a frequency f, the current produced by this rotating charge element is di = dq * f = (q/L)f dl. The magnetic moment (dM) due to this current loop is di * A, where A is the area of the loop formed by the rotation. The area swept by the element at radius l is A = π

Q: What are the options for this CUET UG Physics question?

A. πqfL² · B. πqfL²/3 · C. qfL · D. qfL²

Question

Q16. An insulating rod of length L carries charge q distributed uniformly on it. The rod is pivoted at one end and is rotated at a frequency f about a fixed perpendicular axis. Then the magnetic dipole moment of the system is:
(a) πqfL2
(b) πqfL²/3
(c) qfL
(d) qfL2

FirstInTest · Accepted Answer

Consider an elemental length dl at a distance l from the pivot. The charge on this element is dq = (q/L)dl. When this element rotates at a frequency f, the current produced by this rotating charge element is di = dq * f = (q/L)f dl. The magnetic moment (dM) due to this current loop is di * A, where A is the area of the loop formed by the rotation. The area swept by the element at radius l is A = πl². So, dM = (q/L)f dl * πl² = (πqf/L)l² dl. To find the total magnetic dipole moment (M) of the rod, integrate dM from l = 0 to l = L:M = ∫₀ᴸ (πqf/L)l² dl = (πqf/L) * [l³/3]₀ᴸ M = (πqf/L) * (L³/3) = πqfL²/3.

Answer

πqfL²

Answer

πqfL²/3

Answer

qfL

Answer

qfL²

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