An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is : [2026]

Question

Smart Achievers · Accepted Answer

(1)

${\vec{B}}_{0} = {\vec{B}}_{A B} + {\vec{B}}_{D E} + {\vec{B}}_{B C D}$

$= \frac{μ_{0} i}{4 π r} \hat{i} + \frac{μ_{0} i}{4 π r} \hat{i} - \frac{μ_{0} i}{2 r} \hat{i}$

$= \frac{μ_{0} i}{2 π r} \hat{i} - \frac{μ_{0} i}{2 r} \hat{i}$

$= \frac{μ_{0} i}{2 π r} (1 - π) \hat{i}$

$= - \frac{μ_{0} i}{2 π r} (π - 1) \hat{i}$

Solution

Q.
An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is : [2026]

1 $- \frac{μ_{0}}{2 π} \frac{I}{r} (π - 1) \hat{i}$

2 $\frac{μ_{0}}{2 π} \frac{I}{r} (π - 1) \hat{i}$

3 $- \frac{μ_{0}}{2 π} \frac{I}{r} (π + 1) \hat{i}$

4 $\frac{μ_{0}}{2 π} \frac{I}{r} (π + 1) \hat{i}$

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Solution

Q. An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is : [2026]

1 -μ02πIr(π-1)i^

2 μ02πIr(π-1)i^

3 -μ02πIr(π+1)i^

4 μ02πIr(π+1)i^