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An electron (mass = $9.1 \times {10^{ - 31}}$ $kg$; charge = $1.6 \times {10^{ - 19}}$ $C$) experiences no deflection if subjected to an electric field of $3.2 \times {10^5}$ $V/m$, and a magnetic fields of $2.0 \times {10^{ - 3}} \,Wb/m^2$. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius.......$m$
The magnetic moments associated with two closely wound circular coils $A$ and $B$ of radius $r_A=10 cm$ and $r_B=20 cm$ respectively are equal if: (Where $N _A, I _{ A }$ and $N _B, I _{ B }$ are number of turn and current of $A$ and $B$ respectively)
A wire carrying current $I$ and other carrying $2I$ in the same direction produces a magnetic field $B$ at the mid point. What will be the field when $2I$ wire is switched off
A loop carrying current $I$ lies in the $x$-y plane as shown in the figure. the unit vector $\hat{ k }$ is coming out of the plane of the paper. the magnetic moment of the current loop is :
In the following hexagons, made up of two different material $P$ and $Q,$ current enters and leaves from points $X$ and $Y$ respectively. In which case the magnetic field at its centre is not zero.
A square loop, carrying a steady current $I,$ is placed in a horizontal plane near a long straight conductor carrying a steady current $I_1$ at a distance $d$ from the conductor as shown in figure. The loop will experience
A particle of specific charge (charge/mass) $\alpha$ starts moving from the origin under the action of an electric field $\vec E = {E_0}\hat i$ and magnetic field $\vec B = {B_0}\hat k$. Its velocity at $(x_0 , y_0 , 0)$ is ($(4\hat i + 3\hat j)$ . The value of $x_0$ is:
A galvanmeter has a coil of resistance $200 \Omega$ with a full scale deflection at $20 \mu \mathrm{A}$. The value of resistance to be added to use it as an ammeter of range $(0-20) \mathrm{mA}$ is:
A square loop of area $25\,cm ^2$ has a resistance of $10\,\Omega$. The loop is placed in uniform magnetic field of magnitude $40.0 T$. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in $1.0 sec$, will be $..........\times 10^{-3}$