A uniform beam of positively charged particle | vedclass

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

Question

(A) Let the positively charged beam be at $x = 0$ and the negatively charged beam be at $x = d$. Since the positive charges move in one direction and

Vedclass · Accepted Answer

Vedclass · Answer

(A) Let the positively charged beam be at $x = 0$ and the negatively charged beam be at $x = d$. Since the positive charges move in one direction and the negative charges move in the opposite direction,the currents $I_1$ and $I_2$ produced by these beams are in the same direction. Let the distance from the first beam be $x$. The magnetic field due to the first beam is $B_1 = \frac{\mu_0 I}{2\pi x}$ (directed into the page). The magnetic field due to the second beam at distance $(d - x)$ is $B_2 = \frac{\mu_0 I}{2\pi (d - x)}$ (directed out of the page). The net magnetic field is $B = B_1 - B_2 = \frac{\mu_0 I}{2\pi} \left( \frac{1}{x} - \frac{1}{d - x} ight)$. At the midpoint $x = d/2$,$B = \frac{\mu_0 I}{2\pi} \left( \frac{1}{d/2} - \frac{1}{d/2} ight) = 0$. As $x 	o 0$,$B 	o \infty$,and as $x 	o d$,$B 	o -\infty$. This behavior is correctly represented by the graph in option $A$.

Column $I$	Column $II$
$(A)$ Two parallel wires with current in the same direction,$P$ is the midpoint.	$(p)$ The magnetic fields $(B)$ at $P$ due to the currents in the wires are in the same direction.
$(B)$ Two coaxial circular loops with current in the same direction,$P$ is the midpoint on the axis.	$(q)$ The magnetic fields $(B)$ at $P$ due to the currents in the wires are in opposite directions.
$(C)$ Two coplanar circular loops with current in opposite directions,$P$ is the midpoint.	$(r)$ There is no magnetic field at $P$.
$(D)$ Two concentric coplanar circular loops with current in the same direction,$P$ is the common center.	$(s)$ The wires repel each other.

Similar Questions

An $\alpha$ particle of energy $10 \ eV$ is moving in a circular path in a uniform magnetic field. The energy of a proton moving in the same path and the same magnetic field will be [mass of $\alpha$ particle $= 4 \times$ mass of proton]. (in $eV$)

The energies required to set up in a cube of side $10 \,cm$ $(i)$ a uniform electric field of $10^7 \,Vm^{-1}$ and (ii) a uniform magnetic field of $0.25 \,Wbm^{-2}$ are respectively about $(\mu_0=4 \pi \times 10^{-7} \,Hm^{-1}, \varepsilon_0=8.9 \times 10^{-12} \,Fm^{-1})$

Two concentric loops $A$ and $B$ of same radius $R = 2 \pi \,cm = 2 \pi \times 10^{-2} \,m$ are placed at right angles to each other. If the currents flowing through $A$ and $B$ are $I_A = 3 \,A$ and $I_B = 4 \,A$ respectively, then the net magnetic field at their common centre is:

Pick out the wrong statement.

Vedclass Test Series

Exam Paper Generator

Online Exam Module