Question

A long, straight wire of radius r carries a current i and is placed horizontally in a uniform magnetic field B pointing vertically upward. The current is uniformly distributed over its cross section. (a) At what points will the resultant magnetic field have maximum magnitude? What will be the maximum magnitude? (b) What will be the minimum magnitude of the resultant magnetic field?

Answer 1

Explanation:

(a) To find the maximum  points will the resultant magnetic field have maximum magnitude:

Since the wire concerned carries current, it will also create a magnetic field around it. And it will be optimum for a long, straight wire at the midpoint called P.

Magnetic field created by current transmission wire $=\frac{\mu_{0} \hat{t}}{2 \pi r}$

Total magnetic field $=B+\frac{\mu_{0} i}{2 \pi r}$

(b) To find the minimum magnitude of the resultant magnetic field:

Magnetic field ( B ) = 0

When r is less than magnetic field  

$r<\frac{\mu_{0} i}{2 \pi B}$

Then  

Case i) B = 0

$r=\frac{\mu_{0} i}{2 \pi B}$  

When r is greater than magnetic field  

$r>\frac{\mu_{0} t}{2 \pi B}$

Case ii) Total magnetic field $=B-\frac{\mu_{0} i}{2 \pi r} r$

Answer 2

Explanation:

 To find the maximum  points will the resultant magnetic field have maximum magnitude:

Since the wire concerned carries current, it will also create a magnetic field around it. And it will be optimum for a long, straight wire at the midpoint called P.

Magnetic field created by current transmission wire =\frac{\mu_{0} \hat{t}}{2 \pi r}=2πrμ0t^

Total magnetic field =B+\frac{\mu_{0} i}{2 \pi r}=B+2πrμ0i

(b) To find the minimum magnitude of the resultant magnetic field:

Magnetic field ( B ) = 0

When r is less than magnetic field  

r < \frac{\mu_{0} i}{2 \pi B}r<2πBμ0i

Then  

Case i) B = 0

r=\frac{\mu_{0} i}{2 \pi B}r=2πBμ0i  

When r is greater than magnetic field  

r > \frac{\mu_{0} t}{2 \pi B}r>2πBμ0t

Case ii) Total magnetic field =B-\frac{\mu_{0} i}{2 \pi r} r=B−2πrμ0ir