Circular region of radius $R$ has uniform magnetic field $B = {B_0} + {B_0}t\left( { - \hat k} \right).\,At\,\,t\, = 0\,$ acceleration of charged particle
$a=\frac{q E}{m}=\frac{q}{m} \frac{R^{2} B_{0}}{2 r}=\frac{q B_{0} R^{2}}{2 m r}$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A closely wound solenoid of $2000$ $turns$ and area of cross-section $1.5 \times 10^{-4}\ m^2$ carries a current of $2.0\, A.$ It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{- 2}$ $tesla$ making an angle of $30^o $ with the axis of the solenoid. The torque on the solenoid will be
A coil having $100$ turns, area of $5 \times 10^{-3} \mathrm{~m}^2$, carrying current of $1 \mathrm{~mA}$ is placed in uniform magnetic field of $0.20 \mathrm{~T}$ such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through $90^{\circ}$ is . . . . . . $\mu \mathrm{J}$.
If a proton enters perpendicularly a magnetic field with velocity $v$, then time period of revolution is $T$. If proton enters with velocity $2 v$, then time period will be
A wire in the form of a circular loop of one turn carrying a current produces a magnetic field $B$ at the centre. If the same wire is looped into a coil of two turns and carries the same current, the new value of magnetic induction at the centre is
A uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5$ $Wb/{m^2}$. If a proton having mass $ = 1.7 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C$ moves in this field vertically downwards with energy $5\, MeV$, then the force acting on it will be
A circular conducting loop of radius $R$ carries a current $I.$ Another straight infinite conductor carrying current $I$ passes through the diameter of this loop as shown in the figure. The magnitude of force exerted by the straight conductor on the loop is
$\alpha $ particle, proton and duetron enters in a uniform (transverse) magnetic field $'B'$ with same acceleration potential find ratio of radius of path followed by these particles.
A cylindrical conductor of radius $R$ is carrying a constant current. The plot of the magnitude of the magnetic field, $B$ with the distance $d$, from the centre of the conductor, is correctly represented by the figure$.....$