Moving Charges And Magnetism

An electron is moving with velocity $\left(2 \hat{\mathrm{i}} + 2 \hat{\mathrm{j}}\right)$ m/s in an electric field of intensity $\overrightarrow{\mathrm{E}} = \hat{\mathrm{i}} + 2 \hat{\mathrm{j}} - 8 \hat{\mathrm{k}}$ volt/m and a magnetic field of $\overrightarrow{\mathrm{B}} = \left(2 \hat{\mathrm{j}} + 3 \hat{\mathrm{k}}\right)$ tesla. Find the magnitude of force on the electron.

The ratio of magnetic field and magnetic moment at the centre of a current carrying circular loop is x. When both the current and radius is doubled the ratio will be

• x/8

• x/4

• x/2

• 2x

Current through ABC and A' B'C' is I. What is the magnetic field at P ? BP = PB' = r (Here C' B' PBC are collinear)

• $\mathrm{B} = \frac{1}{4\mathrm{\pi }} \frac{2\mathrm{I}}{\mathrm{r}}$

• $\mathrm{B} = \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }} \left(\frac{2\mathrm{I}}{\mathrm{r}}\right)$

• $\mathrm{B} = \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }} \left(\frac{\mathrm{I}}{\mathrm{r}}\right)$

• Zero

The magnetic field at the point of intersection of diagonals of a square wire loop of side L carrying a current I is

• $\frac{{\mathrm{\mu }}_0\mathrm{I}}{\mathrm{\pi L}}$

• $\frac{2{\mathrm{\mu }}_0\mathrm{I}}{\mathrm{\pi L}}$

• $\frac{\sqrt{2} {\mathrm{\mu }}_0\mathrm{I}}{\mathrm{\pi L}}$

• $\frac{2\sqrt{2} {\mathrm{\mu }}_0\mathrm{I}}{\mathrm{\pi L}}$

A straight wire of length 2 m carries a current of 10 A. If this wire is placed in a uniform magnetic field of 0.15 T making an angle of 45° with the magnetic field, the applied force on the wire will be

• 1.5 N

• 3 N

• $3\sqrt{2} \mathrm{N}$

• $\frac{3}{\sqrt{2}} \mathrm{N}$

A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of 60° and the work done is W. The torque on the magnetic needle at this position is

• $2\sqrt{3} \mathrm{W}$

• $\sqrt{3} \mathrm{W}$

• $\frac{\sqrt{3}}{2} \mathrm{W}$

• $\frac{\sqrt{3}}{4} \mathrm{W}$

When a proton is released from rest in a room, it starts with an initial acceleration a₀ towards west. When it is projected towards north with a speed v₀ it moves with an initial acceleration 3a₀ towards west. The electric and magnetic fields in the room are

• $\frac{{\mathrm{ma}}_0}{\mathrm{e}} \mathrm{west}, \frac{2{\mathrm{ma}}_0}{{\mathrm{ev}}_0} \mathrm{up}$

• $\frac{{\mathrm{ma}}_0}{\mathrm{e}} \mathrm{west}, \frac{2{\mathrm{ma}}_0}{{\mathrm{ev}}_0} \mathrm{down}$

• $\frac{{\mathrm{ma}}_0}{\mathrm{e}} \mathrm{east}, \frac{3{\mathrm{ma}}_0}{{\mathrm{ev}}_0} \mathrm{up}$

• $\frac{{\mathrm{ma}}_0}{\mathrm{e}} \mathrm{east}, \frac{3{\mathrm{ma}}_0}{{\mathrm{ev}}_0} \mathrm{down}$

A current loop in a magnetic field

• experiences a torque whether the field is uniform or non- uniform in all orientations

• can be in equilibrium in one orientations

• can be equilibrium in two orientations, both the equilibrium states are unstable

• can be in equilibrium in two orientations, one stable while the other is unstable

A bar magnet of length l and magnetic dipole moment M is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be

• M

• $\frac{3}{\mathrm{\pi }} \mathrm{M}$

• $\frac{2}{\mathrm{\pi }} \mathrm{M}$

• $\frac{\mathrm{M}}{2}$

A galvanometer having internal resistance 10 Ω requires 0.01 A for a full scale deflection. To convert this galvanometer to a voltmeter of full-scale deflection at 120 V, we need to connect a resistance of

• 11990 Ω in series

• 11990 Ω in parallel

• 12010 Ω in series

• 12010 Ω in parallel