JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • question_answer
    In the figure shown, the magnetic field at the point P is

    A)  \[\frac{2{{\mu }_{0}}i}{3\pi a}\sqrt{4-{{\pi }^{2}}}\]                        

    B)  \[\frac{2{{\mu }_{0}}i}{3\pi a}\sqrt{4+{{\pi }^{2}}}\]

    C)  \[\frac{2{{\mu }_{0}}i}{3\pi a}(4+{{\pi }^{2}})\]                 

    D)  \[\frac{2{{\mu }_{0}}i}{3\pi a}(4-{{\pi }^{2}})\]

    Correct Answer: B

    Solution :

     Idea Magnetic field due to a infinite length wire at its end point is given by\[\frac{{{\mu }_{0}}I}{4\pi d}\]and due to semicircular conducting arc is\[\frac{{{\mu }_{0}}I}{4R}\]at centre. Net magnetic field at point P, \[{{\mathbf{B}}_{P}}={{({{\mathbf{B}}_{1}})}_{P}}+{{({{\mathbf{B}}_{2}})}_{P}}+{{({{\mathbf{B}}_{3}})}_{P}}+{{({{\mathbf{B}}_{4}})}_{P}}+{{({{\mathbf{B}}_{5}})}_{P}}\] where, \[{{({{\mathbf{B}}_{1}})}_{P}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{3a}{2} \right)}(-j)\](semi-infinite wire) \[{{({{\mathbf{B}}_{2}})}_{P}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{3a}{2} \right)}(+\mathbf{k}),{{({{\mathbf{B}}_{3}})}_{P}}=0\] \[{{({{\mathbf{B}}_{4}})}_{P}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{a}{2} \right)}(-\mathbf{k})\] \[{{({{\mathbf{B}}_{5}})}_{P}}=\frac{{{\mu }_{0}}i}{4\pi \left( \frac{a}{2} \right)}(-\mathbf{j})\] \[\therefore \]\[{{\mathbf{B}}_{P}}=\frac{{{\mu }_{0}}i}{2a}\left[ -\left( \frac{1}{3\pi }+\frac{1}{\pi } \right)j-\left( 1-\frac{1}{3} \right)k \right]\] \[{{\mathbf{B}}_{P}}=\frac{{{\mu }_{0}}i}{3a}\left[ \frac{2}{\pi }j+k \right]\] \[\Rightarrow \]\[|{{\mathbf{B}}_{P}}|=\frac{2{{\mu }_{0}}i}{3\pi a}\sqrt{4+{{\pi }^{2}}}\] TEST Edge Questions related to calculation of net magnetic field at a point are asked frequently, to solve such problem it is beneficial to learn some standard result e.g., magnetic field at centre of current carrying circular coil is\[B=\frac{{{\mu }_{0}}I}{2R}\]and magnetic field due to infinitely long current carrying straight wire near its centre is\[B=\frac{{{\mu }_{0}}I}{2\pi d}\]

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