An electron is projected with velocity \$\mathrm{v}_{0}\$ in a uniform electric field \$\mathrm{E}\$ perpendicular to the field. Again it is projected with velocity \$\mathrm{v}_{0}\$ perpendicular to a uniform magnetic field \$\mathrm{B}\$. If \$r_1\$ is initial radius of curvature just after entering in the electric field and \$\mathrm{r}_{2}\$ is initial radius of curvature just after entering in magnetic field then the ratio \$\mathrm{r}_{1}/\mathrm{r}_{2}\$ is equal to :

For electric field,Refer image 1.\$q\rightarrow\$ change of electron.Four due to electric field \$=qE\$\$\therefore \dfrac{mv_{0^2}}{R_e}=qE\$where \$R_e\$ is radius of curvature,\$\therefore \dfrac{1}{R_e}=\dfrac{qE}{mv_{0^2}}\$\$\therefore =\dfrac{mv_{0^2}}{qE}=r_1\$In magnetic field,Focus on electron \$=qV_0B\$Refer image 2.Balancing of centripetal and centrifugal focuses,\$\dfrac{mV_{0^2}}{Rb}=qV_0B\$ \$\therefore R_b=\dfrac{mV_0}{qB}=r_2\$\$\therefore \dfrac{r_1}{r_2}=\dfrac{R_c}{R_b}=\dfrac{\dfrac{mV_{0^2}}{qE}}{\dfrac{mV_0}{qB}}\$\$\therefore \dfrac{r_1}{r_2}=\dfrac{BV_0}{E}\$Answer is option \$D\$

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>>Class 12
>>Physics
>>Moving Charges and Magnetism
>>Magnetic Field and Force
>> $Magnetic Field due to Straight Current C...$

Magnetic Field due to Straight Current Carrying Conductor

Q: A magnetic compass needle is placed in the plane of paper near point A as shown in Figure 13.6. In which plane should a straight current carrying conductor be placed so that it passes through A and there is no change in the deflection of the compass? Under what condition is the deflection maximum and why?

According to the Right hand thumb rule, the direction of the magnetic field is perpendicular to the direct of the current and if a magnetic compass is brought closer to the current carrying conductor, the deflection is maximum. But when the needle is placed near the point A on the plane as that of the paper as given in the figure, there is no deflection. Hence, the current carrying conductor needs to be placed in the same plane as that of the paper, for the magnetic compass to attain no deflection.

Q: A magnetic compass shows a deflection when placed near a current carrying wire. How will the deflection of the compass get affected if the current in the wire is increased? Support your answer with a reason.

It the current is increased in the conductor the deflection of the compass needle increases. This is because the strength of magnetic field varies directly as the magnitude of the electric current, or the current passing through the wire.

Q: On which factors do the magnetic field due to current carrying conductor depend ?

The magnitude of the magnetic field produced at the given point:1. Magnetic field Increases as the current through wire increases.2. Magnetic field Decreases as the distance from the wire increases.

Q: A horizontal wire 0.1 m long carries a current of 5 A. Find the magnitude of the magnetic field, which can support the weight of the wire. Assume wire to be of mass \$3 \times 10^{-3}kg m^{-1}\$ :

Here, \$I=5 A; 1= 0.1 m\$;Mass per unit length of the wire \$\lambda =3 \times 10^{-3} kg m^{-1}\$Therefore, weight of the wire, \$mg=3 \times 10^{-3} \times 0.1 \times 9.8 = 2.94 \times 10^{-3}N\$Let B be the strength of the magnetic field applied. \$F=BIl = B \times 5 \times 0.1 = 0.5 B\$In equilibrium, \$F=mg\$\$\therefore 0.5 B = 2.94 \times 10^{-3}\$\$B \displaystyle =\frac{2.94 \times 10^{-3}}{0.5} = 5.88 \times 10^{-3}T\$

Q: Describe some experiment to show that the magnetic field is associated with an electric current.

We take a thick insulated copper wire and fix it in such a way that the portion \$AB\$ of the wire is in the north-south direction as shown in fig. A plotting compass \$M\$ is placed under the wire \$AB\$. The two ends of the wire are connected to a battery through a switch. When no current is flowing in the wire \$AB\$, the compass needle is parallel to the wire \$AB\$ and points in the usual north-south direction. When current is passed through wire \$AB\$ by closing the switch, we find that the compass needle is deflected from its north-south position. On opening the switch, the compass needle returns to its original position.Thus, the deflection of compass needle by the current carrying wire shows that magnetic field is associated with an electric current.

Q: (i) With the help of an activity, explain the method of inducing electric current in a coil with a moving magnet. State the rule used to find the direction of electric current thus generated in the coil. (ii) Two circular coils P and Q are kept close to each other, of which coil P carries a current. What will you observe in Q? (a) If current in the coil P is changed? (b) If both the coils are moved in the same direction with the same speed? Give reason.

(i) Activity: (a) Take a coil of wire AB having a large number of turns. (b) Connect the ends of the coil to a galvanometer (c) Take a strong bar magnet and move its north pole towards the end B of the coil. (d) There is a momentary deflection in the needle of the galvanometer, say to the right. This indicates the presence of a current in the coil AB. The deflection becomes zero the moment the motion of the magnet stops. (e) Now withdraw the north pole of the magnet away from the coil. Now the galvanometer is deflected toward the left, showing that the current is now set up in the direction opposite to the first. (f) Place the magnet stationary at a point near to the coil, keeping its north pole towards the end B of the coil. We see that the galvanometer needle deflects toward the right when the coil is moved towards the north pole of the magnet. Similarly, the needle moves toward left when the coil is moved away. When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer drops to zero. To find the direction of electric current Fleming's right hand rule is applied. According to it, if we stretch the forefinger, middle finger and thumb of our right hand mutually perpendicular in such a way that thumb points along the direction of motion of conductor, forefinger along the direction of magnetic field; then the middle finger points along the direction of induced current. (ii) (a) When current in P is changed, the field associated with Q will vary causing an induced current in Q. (b) If both the coils are moved in the same direction with same speed, there will not be any change in the field associated with Q. Hence no current will be induced in Q.

Q: Draw the magnetic field lines due to current through circular loop.

Magnetic lines of force are circular near the wire became parallel at the middle point of the coil.

Q: A tesla is equivalent to a

Tesla is the SI unit of magnetic field strength. A particle carrying a charge of coulomb and passing through a magnetic field of 1 tesla at a speed of 1 meter per second perpendicular to said field experiences a force with magnitude 1 newton.\$T=V.S/m^2=N/A.m\$

Q: Describe an activity to demonstrate the direction of the magnetic field generated around a current carrying conductor .

• Take a straight thick copper wire and place it between the points X and Y in an electric circuit. The wire XY is kept perpendicular to the plane of the paper.• Horizontally place a small compass near this copper wire. See the position of its needle.• Pass the current through the circuit by inserting the key into the plug.• Observe the change in the position of the compass needle and thedirection of deflection.•Interchange the battery connection in the circuit so that the direction of the current in the copper wire changes.• Observe the change in the direction of deflection of the needle.