Two short magnets with their axes horizontal | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For a short magnet placed such that its axis is perpendicular to the magnetic meridian,the magnetic field at the center of the needle (on the axia

Vedclass · Accepted Answer

$64:125$

Vedclass · Answer

(C) For a short magnet placed such that its axis is perpendicular to the magnetic meridian,the magnetic field at the center of the needle (on the axial line of the magnet) is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}$.Since the needle remains undeflected,the magnetic fields produced by both magnets at the position of the needle must be equal in magnitude and opposite in direction.Thus,$B_1 = B_2$.$\frac{\mu_0}{4\pi} \cdot \frac{2M_1}{d_1^3} = \frac{\mu_0}{4\pi} \cdot \frac{2M_2}{d_2^3}$.This simplifies to $\frac{M_1}{M_2} = \left( \frac{d_1}{d_2} \right)^3$.Given $d_1 = 40 \, cm$ and $d_2 = 50 \, cm$,we have $\frac{M_1}{M_2} = \left( \frac{40}{50} \right)^3 = \left( \frac{4}{5} \right)^3 = \frac{64}{125}$.

Two short magnets with their axes horizontal and perpendicular to the magnetic meridian are placed with their centres $40 \, cm$ east and $50 \, cm$ west of a magnetic needle. If the needle remains undeflected,the ratio of their magnetic moments $M_1:M_2$ is

Similar Questions

$A$ magnet suspended in a uniform magnetic field is heated so as to reduce its magnetic moment by $19 \%$. By doing this,the time period of the magnet approximately

$A$ planet has a magnetic dipole moment of $27 \times 10^{22} \ A \ m^2$. If the radius of the planet is $300 \ km$, what would be the magnetic field at its equator (in $T$)? (Take $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \ m/A$)

$A$ short bar magnet placed with its axis at $30^{\circ}$ to a uniform external magnetic field of $0.5 \ T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment is . . . . . . .

The pole strength of a bar magnet is $48 \, A \cdot m$ and the distance between its poles is $25 \, cm$. The torque required to hold the magnet at an angle of $30^{\circ}$ with a uniform magnetic field of flux density $0.15 \, N \cdot A^{-1} \cdot m^{-1}$ is . . . . . . $N \cdot m$.

$A$ short bar magnet experiences a torque of magnitude $0.64 \ J$ when it is placed in a uniform magnetic field of $0.32 \ T$. The magnetic moment of the magnet is . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module