Explanation
A proton and alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be:
Current sensitivity
$$I_g=\dfrac{NBA}{K}=\dfrac{5div}{mA}=5000\displaystyle\frac{div}{A}$$
Voltage sensitivity $$v_s=\dfrac{NBA}{KR}=20\dfrac{div}{v}$$
$$\Rightarrow\displaystyle\frac{I_g}{v_s}=R$$
$$\Rightarrow R=\displaystyle\frac{5000}{20}=250\Omega$$
Option 1 is correct.
The work required to rotate a magnetic needle by 60$$^{0}$$ from equilibrium position in a uniform magnetic field is W. The torque required to hold it in that position is
A magnet of moment 4Am$$^{2}$$ is kept suspended in a magnetic field of induction $$5\times 10^{-5}T$$. The workdone in rotating it through 180$$^{0}$$ is
Assertion (A): A magnet remains stable, If it aligns itself with the field
Reason (R): The P.E. of a bar magnet is minimum, if it is parallel to magnetic field.
The ratio of magnetic potential due to magnetic dipole at the end to that at the broad side on a position for the same distance from it is :
Maximum P.E. of magnet of moment M situated in a magnetic field of induction B, is
The dimensional formula for magnetic moment is :
Please disable the adBlock and continue. Thank you.