Question

An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional to-

• A. $\sqrt { \frac { B } { v } }$
• B. $\frac { B } { v }$
• C.
• D.

Answer 1

Answer: (A)

$\sqrt { \frac { B } { v } }$

Answer 2

Magnetic field at the centre of n turn coil carrying current i $B = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \pi n i } { r }$ ......(i)

For single turn n =1

$B = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \pi i } { r }$ .....(ii)

If the same wire is turn again to form a coil of three turns i.e. n = 3 and radius of each turn $r ^ { \prime } = \frac { r } { 3 }$

So new magnetic field at centre $B ^ { \prime } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \pi ( 3 ) } { r ^ { \prime } }$

⇒ $B ^ { \prime } = 9 \times \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \pi i } { r }$ ......(iii)

Comparing equation (ii) and (iii) gives $B ^ { \prime } = 9 B$.