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Need clarity, kindly explain! - Magnetic Effects of Current and Magnetism - JEE Main-4

Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge q is passing through their midpoint P at angle \theta=45^{\circ} with the horizontal line , as shown in figure. what would be the magnetitude of force on the particle at that instant? (d is much large than the dimensions of the dipole)

  • Option 1)

    \left ( \frac{\mu_{0}}{4\pi} \right )\frac{M}{\left ( \frac{d}{2} \right )^{3}}\times qv

  • Option 2)

    0

  • Option 3)

    \sqrt{2}\left ( \frac{\mu_{0}}{4\pi} \right )\frac{M}{\left ( \frac{d}{2} \right )^{3}}\times qv

  • Option 4)

    \left ( \frac{\mu_{0}}{4\pi} \right )\frac{2M}{\left ( \frac{d}{2} \right )^{3}}\times qv

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B_{1}=2\left ( \frac{\mu_{0}}{4\pi} \right )\frac{M}{ \frac{d}{2} }

B_{2}=\left ( \frac{\mu_{0}}{4\pi} \right )\frac{M}{\left ( \frac{d}{2} \right )^{3}}

B_{1}=B_{2}

Net magnetic field will be at 45^{\circ}

The direction of Bnet & velocity of charge is same,

hence charge will experience no force.

 


Option 1)

\left ( \frac{\mu_{0}}{4\pi} \right )\frac{M}{\left ( \frac{d}{2} \right )^{3}}\times qv

Option 2)

0

Option 3)

\sqrt{2}\left ( \frac{\mu_{0}}{4\pi} \right )\frac{M}{\left ( \frac{d}{2} \right )^{3}}\times qv

Option 4)

\left ( \frac{\mu_{0}}{4\pi} \right )\frac{2M}{\left ( \frac{d}{2} \right )^{3}}\times qv

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