Question

A uniform current carrying ring of mass m and radius R is connected by a massless string as shown. A uniform magnetic field B₀ exist in the region to keep the ring in horizontal position, then the current in the ring is (l = length of the string) –

• A. $\frac { \mathrm { mg } } { \pi \mathrm { RB } _ { 0 } }$
• B. $\frac { \mathrm { mg } } { \mathrm { RB } _ { 0 } }$
• C. $\frac { \mathrm { mg } } { 3 \pi \mathrm { RB } _ { 0 } }$
• D.

Answer 1

Answer: (D)

Answer 2

t due to B₀ = B₀ × I × pR²

t due to mg = mg × R

mg R = B₀ I × pR² (for equilibrium)

\ I = $\frac { \mathrm { mg } } { \mathrm { B } _ { 0 } \pi \mathrm { R } ^ { 2 } }$